2004 by kouchmar

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                                                       ‫اﻟﻜﻔﺎءات اﻟﻤﺴﺘﮭﺪﻓﺔ‬

                                                    ‫اﻟﺘﻌﺮف ﻋﻠﻰ داﻟﺔ ﻛﺜﯿﺮ ﺣﺪود و ﻋﻠﻰ‬

                                                                             ‫درﺟﺘﮭﺎ.‬

                                                    ‫ﺤل ﻤﺴﺎﺌل ﺘﺴﺘﺨﺩﻡ ﻓﻴﻬﺎ ﻤﻌﺎﺩﻻﺕ ﺃﻭ‬

                                                           ‫ﻤﺘﺭﺍﺠﺤﺎﺕ ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﯿﺔ‬




       ‫ﻟﻘﺪ ﻗﺪم ﺗﻌﺮﯾﻒ ﺟﺪر ﻛﺜﯿﺮ ﺣﺪود ﻟﯿﺲ‬
‫ﺑﮭﺪف ﺣﻞ اﻟﻤﻌﺎدﻻت ذات درﺟﺔ أﻛﺒﺮ ﻣﻦ‬
                                       ‫ﺛﻼﺛﺔ‬
‫و إﻧﻤﺎ ﻻﺳﺘﻌﻤﺎﻟﮫ ﻓﻲ ﺗﺤﻠﯿﻞ ﻛﺜﯿﺮات اﻟﺤﺪود.‬         ‫‪ v‬ﯾﺘﻢ ﻓﻲ ھﺬا اﻟﻔﺼﻞ اﻟﺮﺑﻂ ﺑﯿﻦ اﻟﺠﺎﻧﺐ‬
    ‫‪ v‬ﯾﺒﻘﻰ ﻣﻔﮭﻮم إﺷﺎرة ﺛﻼﺛﻲ اﻟﺤﺪود ﻣﻦ‬            ‫اﻟﺠﺒﺮي اﻟﻤﺘﻤﺜﻞ ﻓﻲ ﺣﻞ ﻣﻌﺎدﻻت و‬
‫أھﻢ ﻣﻤﯿﺰات ھﺬا اﻟﻔﺼﻞ ﺑﺎﻋﺘﺒﺎره ﺟﺪﯾﺪ‬                                       ‫ﻣﺘﺮاﺟﺤﺎت‬
‫ﻋﻠﻰ اﻟﺘﻼﻣﯿﺬ و ﻧﻈﺮا ﻟﺘﻨﻮع اﺳﺘﻌﻤﺎﻻﺗﮫ‬            ‫و اﻟﺠﺎﻧﺐ اﻟﺒﯿﺎﻧﻲ اﻟﻤﺘﻤﺜﻞ ﻓﻲ دراﺳﺔ اﻟﺪوال.‬
            ‫ﻓﻲ ﻣﺨﺘﻠﻒ اﻟﻔﺼﻮل اﻟﻘﺎدﻣﺔ.‬
      ‫‪ v‬ﯾﺴﻤﺢ ﻣﻦ ﺟﮭﺔ أﺧﺮى ھﺬا اﻟﻔﺼﻞ‬
    ‫ﺑﺈﻋﺎدة اﺳﺘﺜﻤﺎر ﻧﺘﺎﺋﺞ اﻟﻔﺼﻞ اﻷول و‬
       ‫اﻟﻤﺘﻤﺜﻠﺔ ﻓﻲ اﺗﺠﺎه ﺗﻐﯿﺮ داﻟﺔ، اﻟﻘﯿﻢ‬
             ‫اﻟﺤﺪﯾﺔ، اﻟﺪوال اﻟﻤﺮﻓﻘﺔ …‬
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                                           ‫اﻟﻨﺸﺎط 5 :‬                                                       ‫اﻷﻧﺸﻄﺔ‬
‫اﻟﮭﺪف : ﺣﻞ ﺑﯿﺎﻧﯿﺎ ﻣﻌﺎدﻟﺔ ﺑﺎﺳﺘﻌﻤﺎل ﻣﻨﺤﻨﻲ داﻟﺘﯿﻦ ﻣﺮﺟﻌﯿﺘﯿﻦ.‬
                                                                                                                ‫اﻟﻨﺸﺎط 1 :‬
   ‫1( ﻧﻼﺣﻆ أن 0 ﻟﯿﺲ ﺣﻼ ﻟِ )∗( . ﻧﻘﺴﻢ اﻟﻄﺮﻓﯿﻦ ﻋﻠﻰ ‪. x‬‬
                              ‫ـ‬
                                                                                     ‫اﻟﮭﺪف : ﺗﺤﻠﯿﻞ ﻋﺪد ﻃﺒﯿﻌﻲ‬
                            ‫‪y‬‬                        ‫2(‬                                                  ‫1(‬
                            ‫2‬                               ‫1 + ‪( x + 2x + 1)( x + 1) = x + 3x + x + 2x‬‬
                                                               ‫3‬            ‫2‬        ‫5‬      ‫3‬    ‫2‬


                                                                     ‫2( 101×1201 = 121301 .121301 ﻟﯿﺲ أوﻟﯿﺎ.‬
                            ‫1‬

                                                                                                                ‫اﻟﻨﺸﺎط 2 :‬
              ‫2-‬     ‫1-‬     ‫0‬       ‫1‬      ‫2‬      ‫‪x‬‬             ‫اﻟﮭﺪف : ﺣﻞ ﻣﻌﺎدﻻت ﺑﺎﺳﺘﻌﻤﺎل اﻟﻌﺒﺎرة اﻟﻤﻨﺎﺳﺒﺔ ﻟﺪاﻟﺔ.‬
                          ‫1-‬                                                    ‫1( 5 + ‪( x + 1)( x + 5 ) = x 2 + 6x‬‬
                          ‫2-‬
                                                                                    ‫5 + ‪( x + 3)2 − 4 = x 2 + 6x‬‬
                                                           ‫2( }1− ,5−{ = 1‪ ، S‬اﻟﺤﻼن ھﻤﺎ ﻓﺼﻠﺘﺎ ﻧﻘﻄﺘﻲ ﺗﻘﺎﻃﻊ ) ‪(C f‬‬
                                                                                                 ‫ﻣﻊ ﻣﺤﻮر اﻟﻔﻮاﺻﻞ.‬
                                          ‫‪ 1‬‬
                                                            ‫) ‪(C f‬‬   ‫}1− ,4−{ = 4 ‪ ، S‬اﻟﺤﻼن ھﻤﺎ ﻓﺼﻠﺘﺎ ﻧﻘﻄﺘﻲ ﺗﻘﺎﻃﻊ‬
       ‫3( ‪ . S = −1, ‬ﯾﺘﻢ اﻟﺘﺤﻘﻖ ﺑﺎﻟﺘﻌﻮﯾﺾ ﻓﻲ )∗( .‬                               ‫ﻣﻊ اﻟﻤﺴﺘﻘﯿﻢ:ذي اﻟﻤﻌﺎدﻟﺔ: 1 + ‪y = x‬‬
                                          ‫‪ 2‬‬
              ‫اﻷﻋﻤﺎل اﻟﻤﻮﺟﮭﺔ‬                                                                                    ‫اﻟﻨﺸﺎط 3 :‬
                                                                        ‫اﻟﮭﺪف : ﺣﻞ ﺑﯿﺎﻧﯿﺎ ﻣﺘﺮاﺟﺤﺔ ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﯿﺔ.‬
                                                                                     ‫‪r‬‬
    ‫ﻣﺠﻤﻮع و ﺟﺪاء ﺣﻠﻲ ﻣﻌﺎدﻟﺔ ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﯿﺔ:‬                                       ‫1(ﺷﻌﺎع اﻻﻧﺴﺤﺎب ھﻮ )3− ,1( ‪u‬‬
‫اﻟﮭﺪف: اﻟﺘﻌﺮف ﻋﻠﻰ ﺑﻌﺾ ﺗﻄﺒﯿﻘﺎت ﻣﺠﻤﻮع و ﺟﺪاء اﻟﺤﻠﯿﻦ.‬
                                           ‫اﻟﺘﻄﺒﯿﻖ1:‬
                                                                                                ‫{‬
                                                            ‫2( 3 + 1; 3 − 1 = ‪ . S‬ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ ھﻲ ﻓﻮاﺻﻞ‬             ‫}‬
                      ‫ﻣﺜﺎل: 5 = ‪ α‬اﻟﺤﻞ اﻟﺜﺎﻧﻲ ھﻮ 5.0‬                              ‫ﻧﻘﻂ ﺗﻘﺎﻃﻊ ) ‪ ( P‬ﻣﻊ ﻣﺤﻮر اﻟﻔﻮاﺻﻞ.‬
                                           ‫اﻟﺘﻄﺒﯿﻖ2:‬         ‫3( ﺣﻠﻮل اﻟﻤﺘﺮاﺟﺤﺔ ھﻲ ﻓﻮاﺻﻞ ﻧﻘﻂ ) ‪ ( P‬اﻟﺘﻲ ﺗﻘﻊ أﺳﻔﻞ‬
     ‫اﻟﺒﺮھﺎن: ﺑﻔﺮض ‪ a + b = S‬و ‪ ab = P‬ﯾﻜﻮن ﻟﺪﯾﻨﺎ:‬
                                                                      ‫ﻣﺤﻮر اﻟﻔﻮاﺻﻞ و ﻣﻨﮫ: ‪. S = 1 − 3,1 + 3 ‬‬
 ‫‪ b = S − a‬و ‪ a ( S − a ) = P‬أي: 0 = ‪a − Sa + P‬‬
   ‫2‬
                                                                            ‫‪‬‬                ‫‪‬‬
         ‫و ﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ‪ a‬ﺣﻞ ﻟﻠﻤﻌﺎدﻟﺔ 0 = ‪x 2 − Sx + P‬‬                       ‫4( ‪S =  −∞,1 − 3  U 1 + 3, +∞ ‬‬
                                                                                ‫‪‬‬              ‫‪ ‬‬           ‫‪‬‬
            ‫ﻛﺬﻟﻚ ‪ b‬ھﻮ ﺣﻞ ﻟﻠﻤﻌﺎدﻟﺔ 0 = ‪. x 2 − Sx + P‬‬                            ‫ﯾﺘﻢ اﻟﺘﺤﻘﻖ ﺑﻮاﺳﻄﺔ ﺟﺪول ﺑﻌﺪ اﻟﺘﺤﻠﯿﻞ.‬
 ‫ﻋﻜﺴﯿﺎ إذا ﻛﺎن ‪ a‬و ‪ b‬ﺣﻠﯿﻦ ﻟﻠﻤﻌﺎدﻟﺔ 0 = ‪x 2 − Sx + P‬‬
                           ‫ﻓﺈن: ‪ a + b = S‬و ‪. ab = P‬‬                                                            ‫اﻟﻨﺸﺎط 4 :‬
   ‫ﻣﺜﺎل: ﻟﺪﯾﻨﺎ 81 = ‪ a + b‬و 77 = ‪ a . ab‬و ‪ b‬ھﻤﺎ ﺣﻼ‬            ‫اﻟﮭﺪف : اﻟﺘﺒﺮﯾﺮ اﻟﮭﻨﺪﺳﻲ ﻟﺤﻞ ﻣﻌﺎدﻟﺔ ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﯿﺔ.‬
              ‫اﻟﻤﻌﺎدﻟﺔ: 0 = 77 + ‪ x 2 − 18x‬أي 7 و 11.‬                                               ‫2‬
                                                                                          ‫‪ 3‬‬     ‫3‬
                                              ‫اﻟﺘﻄﺒﯿﻖ3:‬                               ‫2( 4 = + 4 + ‪x =  ‬‬
                                         ‫اﻟﺒﺮھﺎن: ﻣﺒﺎﺷﺮ‬                                   ‫‪ 2‬‬     ‫2‬
                                                   ‫ﻣﺜﺎل:‬                                      ‫‪b ‬‬     ‫‪b‬‬
                                                                                                            ‫2‬

                         ‫1‬                                                                ‫3( + ‪x =   + c‬‬
‫‪m‬‬       ‫1- ∞−‬          ‫−‬          ‫0‬        ‫∞+ 1‬                                               ‫‪2‬‬      ‫2‬
                         ‫3‬                                                                  ‫2‬
 ‫∆‬        ‫-‬          ‫-‬        ‫+‬        ‫+‬         ‫+‬                                  ‫‪4‬‬      ‫4‬
                                                                                ‫اﻟﺘﻄﺒﯿﻖ: 5 = + 5 + ‪x =  ‬‬
 ‫‪c‬‬                                                                                  ‫‪2‬‬      ‫2‬
          ‫+‬          ‫+‬        ‫+‬        ‫-‬         ‫+‬
 ‫‪a‬‬                                                                         ‫3‬
                                                                             ‫ﺗﻜﺘﺐ اﻟﻤﻌﺎدﻟﺔ ﻋﻠﻰ اﻟﺸﻜﻞ: 2 ‪x + 10 = x‬‬
  ‫‪b‬‬                                                                        ‫2‬
‫−‬         ‫-‬          ‫+‬        ‫+‬        ‫+‬         ‫-‬
  ‫‪a‬‬                                                                                                     ‫2‬
                                                                                           ‫‪3‬‬       ‫3‬
        ‫ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﺒﺮھﻨﺔ ﯾﺘﻢ اﻻﺳﺘﻨﺘﺎج اﻧﻄﻼﻗﺎ ﻣﻦ اﻟﺠﺪول.‬                               ‫4 = + 01 + ‪x =  ‬‬
                                                                                           ‫‪4‬‬       ‫4‬



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                                                                     ‫اﻟﻤﻌﺪﻻت و اﻟﻤﺘﺮاﺟﺤﺎت ﻣﻀﺎﻋﻔﺔ اﻟﺘﺮﺑﯿﻊ:‬
                         ‫71 1( 1 + ‪. f : x → x 2 + x‬‬               ‫اﻟﮭﺪف: ﺣﻞ ﻣﻌﺎدﻻت و ﻣﺘﺮاﺟﺤﺎت ﻣﻀﺎﻋﻔﺔ اﻟﺘﺮﺑﯿﻊ.‬
                        ‫2( 1 − ‪. f : x → − x 2 + x‬‬                                       ‫{‬           ‫}‬
                                                              ‫1( اﻟﺘﻄﺒﯿﻖ: 2 ,2 − = 1‪S 2 = {−2, −1,1, 2} ، S‬‬
                        ‫3( 1 + ‪. f : x → − x 2 + x‬‬
                                                                                                          ‫∅ = 3‪S‬‬
                                ‫1‬                                   ‫2(دراﺳﺔ اﻟﻤﺜﺎل: ‪S =  −2, − 3  U  3, 2 ‬‬
                                                                        ‫‪‬‬         ‫‪ ‬‬        ‫‪‬‬
                            ‫.‬     ‫81 2( ﺳﺎﺑﻘﺘﺎ ھﻤﺎ: 1 و‬
                                ‫3‬
                                                                        ‫اﻟﺘﻄﺒﯿﻖ: ‪S =  −∞, − 5  U  5, +∞ ‬‬
                                                                              ‫‪‬‬        ‫‪ ‬‬            ‫‪‬‬



                  ‫.‬
                      ‫1‬
                        ‫3( اﻟﻘﯿﻤﺔ اﻟﺤﺪﯾﺔ اﻟﻌﻈﻤﻰ ھﻲ:‬                         ‫ﺗــﻤـــﺎرﯾــــﻦ‬
                      ‫3‬                                                                              ‫1 ﺻﺤﯿﺢ .‬
                ‫1‬
          ‫4( ‪ f‬ﻣﺘﺰاﯾﺪة ﺗﻤﺎﻣﺎ ﻋﻠﻰ اﻟﻤﺠﺎل [ ، ∞-].‬
                ‫3‬                                                                                    ‫2 ﺧﺎﻃﺊ .‬
          ‫1‬
        ‫‪ f‬ﻣﺘﻨﺎﻗﺼﺔ ﺗﻤﺎﻣﺎ ﻋﻠﻰ اﻟﻤﺠﺎل [∞+ ، ].‬                                                          ‫ﺧﺎﻃﺊ .‬
           ‫3‬                                                                                                   ‫3‬
                                                                                                      ‫4 ﺧﺎﻃﺊ.‬
                      ‫42 − ‪. f : x → 3x 2 − 6 x‬‬         ‫91‬
                                                                                                      ‫5 ﺻﺤﯿﺢ.‬
           ‫21 + ‪. P ( x ) = x 3 + 7 x 2 + 16 x‬‬     ‫02 1(‬
                                      ‫درﺟﺘﮫ 3.‬                                                      ‫6 1 ( 0.‬
              ‫5 + ‪P ( x ) = x − 3x − 11x‬‬
                           ‫3‬       ‫2‬
                                                   ‫2(‬                 ‫2( 3( 4( 5( ﻟﯿﺴﺖ دوال ﻛﺜﯿﺮات ﺣﺪود.‬
                                      ‫درﺟﺘﮫ 3.‬
              ‫54 + ‪P ( x ) = x − x − 21x‬‬
                          ‫3‬      ‫2‬
                                                   ‫3(‬                 ‫3( ﺻﺤﯿﺢ.‬      ‫2( ﺧﺎﻃﺊ.‬      ‫7 1( ﺻﺤﯿﺢ.‬
                                      ‫درﺟﺘﮫ 3.‬
                                                                      ‫3( ﺻﺤﯿﺢ.‬      ‫2( ﺧﺎﻃﺊ.‬       ‫8 1( ﺻﺤﯿﺢ.‬
                         ‫41 − ‪. P ( x ) = 12 x‬‬     ‫4(‬
                                                                                    ‫5( ﺧﺎﻃﺊ.‬        ‫4( ﺧﺎﻃﺊ.‬
                                      ‫درﺟﺘﮫ 1.‬
                                                                                                      ‫ﺻﺤﯿﺢ.‬    ‫9‬
           ‫6 − ‪P ( x ) + Q( x ) = − x 2 + 5 x‬‬           ‫12‬
           ‫1( 4 − ‪P ( x ) − Q( x ) = −5x 2 − 3x‬‬                                                           ‫01 2(.‬
           ‫31 − ‪2P ( x ) + 3Q( x ) = 14 x‬‬                                                                ‫11 2(.‬

        ‫2 − ‪P ( x ) + Q( x ) = 2 x 3 − 2 x 2 + x‬‬                                                         ‫21 3(.‬
       ‫9 − ‪. P ( x ) − Q( x ) = 2 x + 2 x + x‬‬
                                 ‫3‬     ‫2‬
                                                         ‫2(‬
                                                                                                         ‫31 1(.‬
        ‫2 + ‪2P ( x ) + 3Q( x ) = 4 x 3 − 6 x 2 + 2 x‬‬
                                                                                                         ‫41 2( .‬
     ‫22 1( درﺟﺔ )‪ P(x‬ھﻲ 5 و ﻣﻌﺎﻣﻞ ﺣﺪه اﻷﻋﻠﻰ 6-‬
    ‫2( درﺟﺔ )‪ Q(x‬ھﻲ 7 و ﻣﻌﺎﻣﻞ ﺣﺪه اﻷﻋﻠﻰ72-‬                                ‫1( ﻷﻧﮭﺎ ﻟﯿﺴﺖ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ‪. ℜ‬‬           ‫51‬
      ‫3( درﺟﺔ )‪ R(x‬ھﻲ4 و ﻣﻌﺎﻣﻞ ﺣﺪه اﻷﻋﻠﻰ 5.‬                               ‫2( ﻷﻧﮭﺎ ﻟﯿﺴﺖ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ‪. ℜ‬‬
                                                                                ‫3( ﻷﻧﮭﺎ ﻟﯿﺴﺖ ﻣﻦ اﻟﺸﻜﻞ‬
            ‫1( 0=)1-(‪ f‬إذن 1- ﺟﺪر ﻟـ )‪.f(x‬‬              ‫32‬             ‫1 + ..... + 1− ‪x → an x n + an −1 x n‬‬
                       ‫ﻧﻔﺲ اﻟﺸﺊ ﻣﻊ 2( و 3(.‬
                                                                            ‫4( 4( ﻷﻧﮭﺎ ﻟﯿﺴﺖ ﻣﻦ اﻟﺸﻜﻞ‬
                     ‫42 1( 4-=‪a=1 , b=0 , c‬‬                            ‫1 + ..... + 1− ‪x → an x n + an −1 x n‬‬
                  ‫2( )2+‪P(x)=(x-1)(x-2)(x‬‬
                   ‫3( اﻟﺠﺬور ھﻲ: 2- ، 2 ، 1.‬                                                             ‫61 1(.‬

‫3‬
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                                ‫1‬         ‫1‬
                       ‫= ' '‪x' = − , x‬‬                                               ‫52 1( 0=)2-(‪P‬‬
                                ‫3‬         ‫6( 2‬
                                                                                           ‫3‬
                       ‫0 = 1 − ‪6x − x‬‬
                            ‫2‬
                                                                  ‫2( 2) − ‪P ( x ) = 4( x + 2 )( x‬‬
                                        ‫3‬                                                  ‫2‬
                      ‫− = ' '‪x ' = 0, x‬‬                                     ‫3‬
                                        ‫7( 2‬                                   ‫3( اﻟﺠﺬور ھﻲ: 2- ،‬
                                                                            ‫2‬
                      ‫0 = ‪2 x + 3x‬‬
                          ‫2‬
                                                                                  ‫12‬
                                    ‫2‬                                          ‫.‬     ‫62 =‪b=5 , a‬‬
                      ‫= ' '‪x' = x‬‬                                                 ‫2‬
                                    ‫3‬       ‫8(‬
                      ‫0 = 4 + ‪9 x − 12 x‬‬
                          ‫2‬
                                                                       ‫72 1=‪. a=-1 , b= 3 , c‬‬
                               ‫03 1( ﺣﻠﯿﻦ: 0 ، 3 .‬                              ‫82 1( 1-2)3-‪f(x)=(x‬‬
                              ‫2( ﺣﻠﺒﻦ: -2 ، 2.‬                        ‫ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ ھﻲ: 4 ، 2.‬
                              ‫3( .ﺣﻠﯿﻦ: -1 ،1.‬                                       ‫1‬        ‫52‬
                          ‫7‬         ‫7‬                               ‫− 2) + ‪f ( x ) = ( x‬‬           ‫2(‬
                             ‫−,‬        ‫4( ﺣﻠﺒﻦ:‬                                      ‫2‬         ‫4‬
                         ‫2‬         ‫2‬                                   ‫ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ ھﻲ: 3- ، 2‬
                               ‫5( ﻻ ﺑﻮﺟﺪ ﺣﻠﻮل.‬                                ‫‪‬‬      ‫3‬       ‫‪11‬‬
                          ‫6( ﺣﻞ ﻣﻀﺎﻋﻒ: -1.‬                     ‫3( ‪f ( x ) = − ( x − ) 2 + ‬‬
                                                                              ‫‪‬‬      ‫2‬        ‫‪4‬‬
                            ‫7( ﺣﻞ ﻣﻀﺎﻋﻒ: 3.‬
                                                                             ‫اﻟﻤﻌﺎدﻟﺔ ﻻ ﺗﻘﺒﻞ ﺣﻠﻮل.‬
                                ‫8( ﺣﻠﺒﻦ: 5 ، 1.‬
                                                                              ‫‪‬‬      ‫7‬        ‫‪25 ‬‬
                          ‫2‬                                     ‫4( ‪f ( x ) = 3( x − ) 2 − ‬‬
                       ‫.‬        ‫9( ﺣﻞ ﻣﻀﺎﻋﻒ:‬                                  ‫‪‬‬      ‫6‬       ‫‪36 ‬‬
                         ‫2‬
                                                                        ‫1‬
                             ‫5‬
                           ‫01( ﺣﻠﺒﻦ : 1 ، ،‬                                 ‫ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ ھﻲ: 2 ،‬
                                                                        ‫3‬
                             ‫7‬
                                                                                 ‫‪‬‬             ‫‪1‬‬
                                                                      ‫5( ‪f ( x ) = ( x − 1) 2 − ‬‬
                            ‫13 ﻣﻤﯿﺰ اﻟﻤﻌﺎدﻟﺔ ﻣﻌﺪوم.‬                              ‫‪‬‬             ‫‪5‬‬
                                                                    ‫1‬            ‫1‬
    ‫23 ﺑﻤﺎ أن ‪ a , b‬ﻣﺘﻌﺎﻛﺴﯿﻦ ﻓﻲ اﻹﺷﺎرة ﻓﺈن اﻟﻤﻌﺎدﻟﺔ‬           ‫−1‬          ‫+1 ,‬       ‫ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ:‬
                                                                     ‫5‬             ‫5‬
                             ‫ﺗﻘﺒﻞ ﺣﻠﺒﻦ ﻣﺘﻤﺎﯾﺰ ﺑﻦ.‬
                                                                                ‫‪‬‬      ‫3‬        ‫‪9‬‬
                                                                ‫6( ‪f ( x ) = −5( x − )2 − ‬‬
                      ‫2 = ' ' ‪x' = 1 , x‬‬                 ‫33‬                     ‫‪‬‬      ‫2‬        ‫‪4‬‬
                                                    ‫1(‬                    ‫ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ ھﻲ: 0 ، 3؟‬
                      ‫) 2 − ‪f ( x ) = ( x − 1)( x‬‬
                          ‫4‬                                                ‫3 = ' '‪x' = 2 , x‬‬
                  ‫2 = ' '‪x' = , x‬‬                                                               ‫92 1(‬
                          ‫3‬
                                                    ‫2(‬                     ‫0 = 6 + ‪x − 5x‬‬
                                                                             ‫2‬
                                       ‫4‬
                  ‫) 2 − ‪f ( x ) = 3( x − )( x‬‬                                                ‫1‬
                                       ‫3‬                                 ‫= ' '‪x' = −3 , x‬‬
                                                                                             ‫2( 2‬
                        ‫1‬             ‫2‬
              ‫− = ' '‪x' = , x‬‬                                            ‫0 = 3 − ‪2x + 5x‬‬
                                                                             ‫2‬
                        ‫3‬             ‫9‬
                                                    ‫3(‬                        ‫3 = ' '‪x ' = 0 , x‬‬
                                      ‫1‬       ‫2‬                                                  ‫3(‬
              ‫) + ‪f ( x ) = −9( x − )( x‬‬                                      ‫0 = ‪x 2 − 3x‬‬
                                      ‫3‬       ‫9‬
                         ‫3‬                                                ‫2− = ' ' ‪x ' = x‬‬
                ‫1 = ' '‪x' = , x‬‬                                                                  ‫4(‬
                         ‫5‬                                                ‫0 = 4 + ‪x 2 + 4x‬‬
                                                    ‫4(‬
                                        ‫3‬                                  ‫1− = ' '‪x' = 5 , x‬‬
                ‫)1 − ‪f ( x ) = −5( x − )( x‬‬                                                      ‫5(‬
                                        ‫5‬                                  ‫0 = 5 − ‪x 2 − 4x‬‬




‫4‬
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                                        ‫9=∆‬                             ‫3 − 9 − '' 3 − 9‬
                                                                 ‫= '‪x‬‬           ‫= ‪,x‬‬
                                        ‫1− = ′ ‪x‬‬                           ‫4‬            ‫4‬            ‫5(‬
                                        ‫‪3−m‬‬                                       ‫3 −9‬        ‫3 +9‬
                                        ‫= ′′ ‪x‬‬                   ‫− ‪f ( x ) = 2( x‬‬      ‫+ ‪)( x‬‬      ‫)‬
                                          ‫‪m‬‬                                         ‫4‬           ‫4‬
            ‫3( ﻟﻤﺎ 1- =‪ m‬اﻟﻤﻌﺎدﻟﺔ ﺗﻘﺒﻞ ﺣﻞ وﺣﯿﺪ3‬
                             ‫‪‬‬     ‫‪7 7‬‬                                            ‫ﺣﻠﺒﻦ: 5- ، 2.‬       ‫1(‬     ‫43‬
               ‫−‪∆ < 0 m ∈ ‬‬          ‫,‬     ‫. ﻟﻤﺎ ‪‬‬                                   ‫2‬
                             ‫‪‬‬     ‫‪8 8‬‬                                            ‫ﺣﻠﺒﻦ: 1 ، .‬         ‫2(‬
                                                                                     ‫3‬
                               ‫اﻟﻤﻌﺎدﻟﺔ ﻻ ﺗﻘﺒﻞ ﺣﻠﻮل.‬
                                                                                    ‫ﻻ ﯾﻮﺟﺪ ﺣﻠﻮل.‬       ‫3(‬
                                              ‫ﻟﻤﺎ‬
                                                                        ‫5 −5−‬   ‫5 +5−‬
                        ‫‪‬‬           ‫7 ‪7 ‬‬             ‫‪‬‬                      ‫,‬             ‫ﺣﻠﺒﻦ:‬      ‫4(‬
            ‫−, ∞ − ‪∆ > 0 m ∈ ‬‬        ‫‪∪‬‬         ‫‪, + ∞‬‬                  ‫2‬          ‫2‬
                        ‫‪‬‬           ‫8 ‪8 ‬‬             ‫‪‬‬                         ‫ﺣﻠﺒﻦ: 91 ، 2-.‬        ‫5(‬
                             ‫اﻟﻤﻌﺎدﻟﺔ ﺗﻘﺒﻞ ﺣﻠﯿﻦ ﻣﺘﻤﺎﯾﺰﯾﻦ.‬

                                                                                     ‫53 1( ) ‪∆ = 4( b′2 − ac‬‬
                        ‫−=‪m‬‬
                                 ‫7‬
                                     ‫= ‪ m‬أو‬
                                                ‫7‬
                                                     ‫ﻟﻤﺎ‬                                ‫2( ‪∆′ = b′2 − ac‬‬
                                 ‫8‬              ‫8‬                     ‫3( إدا ﻛﺎن 0≥/ ∆ ﻓﺈن: 0≥ ∆ و ﻣﻨﮫ‬
                           ‫اﻟﻤﻌﺎدﻟﺔ ﺗﻘﺒﻞ ﺣﻞ ﻣﻀﺎﻋﻒ.‬                       ‫اﻟﻤﻌﺎدﻟﺔ )‪ (E‬ﺗﻘﺒﻞ ﺣﻠﯿﻦ ﻣﺘﻤﺎﯾﺰﯾﻦ ھﻤﺎ:‬
            ‫4( ﻟﻤﺎ 3 =‪ m‬اﻟﻤﻌﺎدﻟﺔ ﺗﻘﺒﻞ ﺣﻞ وﺣﯿﺪ 1-.‬                           ‫∆ −‪−b‬‬                 ‫∆ +‪−b‬‬
                                        ‫ﻟﻤﺎ 3 ≠‪m‬‬                     ‫= ′‪x‬‬               ‫= ′′ ‪, x‬‬
                                                                                ‫‪2a‬‬                    ‫‪2a‬‬
                                     ‫52 = ∆‬
                                                                                                        ‫و ﻣﻨﮫ:‬
                                     ‫1− = ′ ‪x‬‬
                                                                          ‫′∆ − ′‪− b‬‬              ‫′∆ + ′‪− b‬‬
                                            ‫‪2+m‬‬                    ‫= ′‪x‬‬                ‫= ′′ ‪, x‬‬
                                     ‫= ′′ ‪x‬‬                                    ‫‪a‬‬                      ‫‪a‬‬
                                            ‫‪3−m‬‬
          ‫=‪ m‬اﻟﻤﻌﺎدﻟﺔ ﺗﻘﺒﻞ ﺣﻞ وﺣﯿﺪ 1-.‬
                                             ‫1‬
                                                ‫5( ﻟﻤﺎ‬                  ‫63 1( 001=’∆ ، 1-=’’‪x’=19 , x‬‬
                                             ‫2‬                          ‫2( 1=’∆ , 99-=’’‪x’=-101 , x‬‬
                                              ‫1‬                                          ‫6‬
                                      ‫≠‪m‬‬         ‫ﻟﻤﺎ‬                        ‫= ′′‪x ′ = x‬‬    ‫3( 0 = '∆ ,‬
                                              ‫2‬                                         ‫2‬
                       ‫1 = ′∆‬
                                             ‫1 + ‪2m‬‬                 ‫3 = ′′ ‪∆ = 1 , t ′ = 2 , t‬‬         ‫73 1(‬
                       ‫= ′′ ‪x ′ = −1 , x‬‬
                                             ‫‪1 − 2m‬‬                 ‫71− = ′′‪∆′ = 81 , u′ = 1 , u‬‬       ‫2(‬

                               ‫04 إﺳﺘﺨﺪام اﻟﺤﺎﺳﺒﺔ اﻟﺒﯿﺎﻧﯿﺔ.‬         ‫3 ( 2 = ′′ ‪∆ = ( 3 − 2 )2 , x ′ = 3 , x‬‬
                                                                    ‫3− = ∆‬                              ‫4(‬
                               ‫14 إﺳﺘﺨﺪام اﻟﺤﺎﺳﺒﺔ اﻟﺒﯿﺎﻧﯿﺔ.‬
                                                                                      ‫83 1( }2,2-{-‪m∈ℜ‬‬
                         ‫(‬         ‫)‬
                           ‫24 1( 3 2 − 4 = 1 − 3‬
                                    ‫2‬
                                                                                          ‫2‬
                                                                                     ‫2( 1=‪x = − .m‬‬
                                           ‫2(‬                                             ‫3‬
                         ‫3 2 − 4 = ′∆‬
                                   ‫3‬          ‫1‬                                    ‫5 + 2 ‪∆′ = m‬‬               ‫93‬
                         ‫= ′‪x‬‬        ‫= ′′ ‪, x‬‬
                                  ‫2‬           ‫2‬                                    ‫1( 5 + ‪x ′ = m − m‬‬
                                                                                                   ‫2‬


    ‫2 ‪(x − x1 )(x − x2 ) = x 2 − ( x1 + x2 )x + x1x‬‬         ‫34‬                   ‫5 + 2 ‪x ′′ = m + m‬‬
                                   ‫ﻣﻤﺎ ﺳﺒﻖ ﻧﻼﺣﻆ أن:‬                 ‫2( ﻟﻤﺎ 0=‪ m‬اﻟﻤﻌﺎدﻟﺔ ﺗﻘﺒﻞ ﺣﻞ وﺣﯿﺪ 1-.‬
                                    ‫(‬
                         ‫3 − ‪f( x ) = x − 2 x‬‬    ‫()‬    ‫)‬                                      ‫( ﻟﻤﺎ 0≠‪m‬‬


‫5‬
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                    ‫0 = 72 − ‪x 2 + 3x‬‬    ‫1(‬             ‫و ﻣﻨﮫ ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ ھﻲ: ) 3 ( , ) 2 (‬
                          ‫01‬                                                   ‫3( ﻧﻔﺲ اﻟﺤﻠﻮل.‬
                    ‫0 = 1+ ‪x2 − x‬‬        ‫2(‬                                                  ‫.‬
                           ‫3‬
                    ‫0 = ‪x 2 − 3x‬‬         ‫3(‬                        ‫44 ) ‪8 x = ( x + 5)(12 − x‬‬
                                                                       ‫2‬



                          ‫1‬     ‫2‬                                  ‫0 = 06 − ‪9 x 2 − 7 x‬‬
                    ‫0 = − ‪x2 − x‬‬         ‫4(‬
                          ‫3‬     ‫3‬                                                     ‫02‬
                                                                   ‫− = ′′ ‪x ′ = 3 , x‬‬
                    ‫5 ( 0 = 2‪x 2 − 2x + 1 − m‬‬                                         ‫9‬
                                                                  ‫ﻃﻮل ﺿﻠﻊ اﻟﻤﺮﺑﻊ ھﻮ: ‪3 m‬‬
                    ‫0 = 32 + ‪x 2 − 10 x‬‬   ‫6(‬
                                                            ‫2) ‪8πr 2 = π( 2 + r‬‬                 ‫54‬
              ‫0 = 4 + ‪x − 7x‬‬
               ‫2‬                                ‫15‬
                                                            ‫0 = 2 − ‪r 2 − 2r‬‬
              ‫33 = ∆‬
                                                            ‫3 + 1 = ′′ ‪r ′ = 1 − 3 , r‬‬
                   ‫33 − 7‬        ‫33 + 7‬
              ‫= ′‪x‬‬        ‫= ′‪, x‬‬                              ‫و ﻣﻨﮫ ﻧﺼﻒ اﻟﻘﻄﺮ ھﻮ 3 + 1 .‬
                      ‫2‬             ‫2‬   ‫25‬
          ‫4 = ‪a + b‬‬                                                                            ‫64‬
          ‫‪‬‬                                                          ‫05 = ‪3 x 2 + 5 x‬‬
          ‫1− = ‪a × b‬‬
              ‫({‬               ‫()‬
         ‫5 − 2, 5 + 2 , 5 + 2, 5 − 2 = ‪S‬‬         ‫})‬                  ‫0 = 05 − ‪3 x 2 + 5 x‬‬
                                                                                        ‫01‬
                                                                      ‫= ‪x = −5 , x‬‬
     ‫52− = ‪a + b‬‬                                                                        ‫3‬
     ‫‪‬‬                                                         ‫01‬
                                                                  ‫و ﻣﻨﮫ ﻃﻮل ﺿﻠﻊ اﻟﻤﺜﻠﺚ ھﻮ:‬
     ‫001 = ‪a × b‬‬                                               ‫3‬
     ‫}) 02−,5 −( ,)5−,02 −({ = ‪S‬‬                                                                 ‫74‬
     ‫41 = ‪a + b‬‬
     ‫‪‬‬
     ‫33 = ‪a × b‬‬
     ‫})3,11( ,)11,3({ = ‪S‬‬
    ‫3 + 1 = ‪a + b‬‬
    ‫‪‬‬
    ‫‪‬‬             ‫3‬                                     ‫84 اﻟﻤﻌﺎدﻻت 1( ، 3( ، 4( ، 5( ﺗﻘﺒﻞ ﺣﻠﯿﻦ‬
    ‫+ 1 = ‪a × b‬‬
    ‫‪‬‬            ‫2‬                                            ‫ﻷن ‪ a , b‬ﻣﺘﻌﺎﻛﺴﯿﻦ ﻓﻲ اﻹﺷﺎرة.‬
        ‫‪‬‬                 ‫‪‬‬                             ‫أﻣﺎ اﻟﻤﻌﺎدﻟﺘﯿﻦ 2( ، 6( ﻓﺎﻟﻤﻤﯿﺰ ﻣﻮﺟﺐ و‬
        ‫‪ 1 + 3 1 + 3 ‬‬
    ‫‪S = ‬‬        ‫,‬      ‫‪‬‬                                                 ‫ﺑﺎﻟﺘﺎﻟﻲ ﺗﻘﺒﻼن ﺣﻠﯿﻦ.‬
        ‫2 ‪‬‬
        ‫‪‬‬            ‫‪2 ‬‬
                         ‫‪‬‬                                ‫ﻣﺠﻤﻮع و ﺟﺪاء اﻟﺤﻠﯿﻦ ﻟﻠﻤﻌﺎدﻟﺔ اﻷوﻟﻰ‬
          ‫0 = ‪a + b‬‬                                                   ‫‪b 3 c‬‬
          ‫‪‬‬                                                         ‫ھﻮ: 2− = , = −‬
          ‫‪‬‬         ‫94 −‬                                               ‫‪a 2 a‬‬
          ‫4 = ‪a × b‬‬
          ‫‪‬‬
                                                            ‫ﻧﻔﺲ اﻟﺸﺊ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﺎدﻻت اﻷﺧﺮى.‬
                                                                                                ‫94‬
               ‫‪ 7 7   7 7 ‬‬                                      ‫ﻧﻘﻮم ﺑﺤﻞ اﻟﻤﻌﺎدﻟﺔ )’‪: (E‬‬
           ‫‪S =  ,− ,  − , ‬‬
               ‫‪ 2 2   2 2 ‬‬                                         ‫)′′ ‪∆ = (x ′ − x‬‬
                                                                                         ‫2‬


‫‪‬‬        ‫01‬                                                              ‫′′ ‪x1 = x ′ , x2 = x‬‬
‫12 = ‪a + b‬‬
‫‪‬‬
‫‪‬‬                                                                       ‫إذن اﻟﻤﻌﺎدﻟﺘﯿﻦ ﻣﺘﻜﺎﻓﺌﺘﯿﻦ.‬
‫1 = ‪a × b‬‬
‫‪‬‬
‫‪‬‬        ‫12‬                                                                                      ‫05‬
    ‫‪‬‬
    ‫‪ 5 − 4 6 5 + 4 6   5 + 4 6 5 − 4 6 ‬‬‫‪‬‬
‫‪S = ‬‬                 ‫‪‬‬
      ‫‪ 21 , 21 ,  21 , 21 ‬‬
                                           ‫‪‬‬
    ‫‪‬‬
    ‫‪‬‬                  ‫‪‬‬                  ‫‪‬‬‫‪‬‬
                                            ‫35‬
‫6‬
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                   c
                     =−
                        5
                                                (1            a − b = 4
                                                              
                   a    3                                     a × b = −1
                               34
                   x ′2 + x ′′2 =
                                9
                                              (2                   {(                 )(
                                                              S = 2 − 3 ,−2 − 3 , 2 + 3 ,−2 + 3             )}
                                                           a − b = 5
                1 1             2                          
                   +       =−                              a × b = 8
               x ′ x ′′         5                               5 − 657 − 5 − 657   5 + 657 − 5 + 657 
                                                                                                         
                                                           S =                    , 
                                                                                                          
               (x ′ − x ′′)2 =  64                                       ,                     ,
                                                               
                                                                   2         2    
                                                                                         2         2    
                                                                                                           
                                 9
                                691
               x ′4 + x ′′4 =                                   a + 3b = 8
                                81                              
                     ′            2m                 58        a × b = 5
                    x + x ′′ = 3
                                                                    5         
                                      :‫1( ﻟﺪﯾﻨﺎ‬                S =  3, , (5,1)
                    x ′ × x ′′ =  1                                 3         
                    
                                  3                                 a − 3b = 7
                                                                     
                                      ′′ m                          a × b = −5
                                     x =
              m=2 , m=-2:‫ و ﻣﻨﮫ‬              6                                           
                                                                                  5
                                     x ′′ 2 = 9                        S =  2,− , (5,−1)
                                                                               2        
                  ′           1− m
                 x + x ′′ = 4
                                                                                                                      .
                 
                                     :‫2( ﻟﺪﯾﻨﺎ‬                                                                  54
                 x ′ × x ′′ = m                               a + b = 8
                                                               
                 
                              2                               1 1 8
                                                                a × b = 15
                                                               
                    ′′        m
    m=0           2 x = − 4
                                                                   
                                                                                   113                     
           :‫ و ﻣﻨﮫ‬                                           S =  4 −
                                                                           113
                                                                               ,4 +     ,  4 + 113 ,4 − 113 
                                                                                                               
    m = 34        x ′′2 + 1 x ′′ − m = 0                          
                                                                          8        8 
                                                                                                8        8 
                                                                                                              
                   
                            4       2

                                                                           x×y =
                                                                                   (x + y )3 − x 3 − y 3         55
          m′ = 1 − 5 ,              m′′ = 1 + 5 (1 59                                      3( x + y )      (1
                             17                                          x × y = 72
         .‫ ﻻ ﯾﻮﺟﺪ ﺣﻠﻮل‬m ∈  ,+∞  ‫2( ﻟﻤﺎ‬
                              12                                          x 2 + y 2 = (x + y ) − 2 xy
                                                                                                 2


               ]              [
                                 17                                                                      (2
          m ∈ − ∞,− 2 ∪  2,  ‫ﻟﻤﺎ‬                                          x 2 + y 2 = 145
                                 12 
                        .‫ﯾﻮﺟﺪ ﺣﻠﯿﻦ ﻣﻮﺟﺒﯿﻦ‬
                                    ]       [                                 c                                  56
            ‫ ﯾﻮﺟﺪ ﺣﻠﯿﻦ‬m ∈ − 2, 2 ‫ﻟﻤﺎ‬                                            = −34                      (1
                                                                              a
                         .‫ﻣﺨﺘﻠﻔﯿﻦ ﻓﻲ اﻹﺷﺎرة‬                                        7      1
                                     17                                       x2 +    x−    =0             (2
             ،‫ ﯾﻮﺟﺪ ﺣﻞ ﻣﻀﺎﻋﻒ‬m =           ‫ﻟﻤﺎ‬                                      34    34
                                     12
           ‫ ﯾﻮﺟﺪ‬m = − 2 ‫ أو‬m = 2 ‫ﻟﻤﺎ‬                                                                             57
                      .‫ﺣﻞ ﻣﻮﺟﺐ و ﺣﻞ ﻣﻌﺪوم‬
                                                     60
                                        1
                             m ∈  − ∞ ,  (1
                                        5
                      m ∈ ]− ∞,−1[ ∪ ]− 1,1[ (2
                               m ∈ ]− 2,3[ (3

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                            1                                            1 1      
P ( x ) > 0 , x ∈  − ∞,−  ∪ ]2,+∞[ ‫1( ﻟﻤﺎ‬                       m ∈  − ∞,  ∪  ,+∞  (4
                            2                                            3  2     
                                  1 
              P ( x ) < 0 , x ∈  − ,2 ‫ﻟﻤﺎ‬                                                             61
                                  2                                            4 97 
                                                                          m ∈  ,  (1
                                 1                                               3 12 
             P ( x ) = 0 , x = − , x = 2 ‫ﻟﻤﺎ‬
                                 2                                      3 + 2 6        
                              3                          m ∈ ]− ∞,−1[ ∪            ,+∞  (2
           P ( x ) = 0 , x = , x = −1 ‫2( ﻟﻤﺎ‬                             5             
                              2
                                                         2 + 6 5 4  2 − 6 5 
                                   3                 m∈        ,−  ∪ 1,
    P ( x ) < 0 , x ∈ ]− ∞,−1[ ∪  ,+∞  ‫ﻟﻤﺎ‬                                             (3
                                   2                    − 11     3          − 11 
                                         1                             .m ‫4( ﻻ ﯾﻮﺟﺪ ﻗﯿﻢ ﻟـ‬
                  P ( x ) > 0 , x ∈  − 1,  ‫ﻟﻤﺎ‬
                                         2
                                                                          x ′ + x ′′ = 23         62
                                                                          
     P ( x ) = 0 , x = −2 , x = 1 , x = 3 ‫3( ﻟﻤﺎ‬                          x ′ × x′′ = 28
      P ( x ) < 0 , x ∈ ]− ∞,−2[ ∪ ]1,3[ ‫ﻟﻤﺎ‬                              x 2 − 23x + 28 = 0
      P ( x ) > 0 , x ∈ ]− 2,1[ ∪ ]3,+∞[ ‫ﻟﻤﺎ‬                              x ′ ≈ 1,28 , x ′′ ≈ 21,7
        P ( x ) = 0 , x = − 3 , x = 3 ‫4( ﻟﻤﺎ‬
                        ]
     P ( x ) > 0 , x ∈ − ∞,− 3 ∪  [ ] 3,+∞[ ‫ﻟﻤﺎ‬           2( x ′ + x ′′ ) = 12
                                                          
                                                                                                   63
                                                          2 x ′ × x ′′ = 9
                P( x ) < 0  , x ∈ ]− 3 , 3 [ ‫ﻟﻤﺎ‬          2 x 2 − 12 x + 9 = 0                     (1
              P ( x ) = 0 , x = −1 , x = 1    ‫5( ﻟﻤﺎ‬      x′ = 3 − 3 /   2 , x ′′ = 3 + 3 /   2
     P ( x ) > 0 , x ∈ ]− ∞,−1[ ∪ ]1,+∞[      ‫ﻟﻤﺎ‬
                  P ( x ) < 0 , x ∈ ]− 1,1[   ‫ﻟﻤﺎ‬            2( x ′ + x ′′ ) = 12
                                                             
                                                             2 x ′ × x ′′ > 9
                                         7
              P( x ) = 0 , x = 0 , x =        ‫6( ﻟﻤﺎ‬
                                         3
                                                             − 2 x 2 + 12 x − 9 > 0                (2
                                  7      
    P ( x ) > 0 , x ∈ ]− ∞,0[ ∪  ,+∞        ‫ﻟﻤﺎ‬                  ]
                                                             x ′′ ∈ 3 − 3 /   2 ,3 + 3 /      2[
                                  3      
                                     7                     X ′ ∈ ]3 − 3 /   2 ,3 + 3 /      2[
                 P ( x ) < 0 , x ∈  0,      ‫ﻟﻤﺎ‬
                                     3                 .‫3( ﺗﺼﺤﯿﺢ: اﻟﻤﺴﺘﻄﯿﻞ ﻟﮫ ﻧﻔﺲ ﻣﺤﯿﻂ اﻟﻤﺮﺑﻊ‬
                                         3                             2( x ′ + x ′′ ) = 2m
                       P( x ) = 0 , x =       ‫7( ﻟﻤﺎ‬                   
                                         2                                           1 2
                                  3                                  x ′ × x′′ = 3 m
                                                                       
               P ( x ) < 0 , x ∈  ,+∞       ‫ﻟﻤﺎ‬
                                  2                                                   1
                                                                       x 2 − 2mx + m 2 = 0
                                       3                                              3
              P ( x ) > 0 , x ∈  − ∞,       ‫ﻟﻤﺎ‬
                                       2                                               8
                                                                               2m − m
                                                                       x ′′ =            3
                              3                 65
           P( x ) = 0 , x =     , x = −2 ‫1( ﻟﻤﺎ‬                                       2
                              2                                                          8
                                     3                                       2m − m
              P ( x ) < 0 , x ∈  − 2,  ‫ﻟﻤﺎ‬                            X′ =             3
                                     2                                              2
                                      3     
       P ( x ) > 0 , x ∈ ]− ∞,−2[ ∪  ,+∞  ‫ﻟﻤﺎ‬
                                      2                                                               64


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                                 2                                                  2
       P ( x ) > 0 , x ∈  − ∞,  ∪ ]1,2[ ‫ﻟﻤﺎ‬                     P( x ) = 0 , x =      , x = 2 ‫2( ﻟﻤﺎ‬
                                 3                                                  3
                                             ‫5( ﻟﻤﺎ‬                                      2 
                                                                      P ( x ) > 0 , x ∈  ,2 ‫ﻟﻤﺎ‬
       P ( x ) = 0 , x = −1 , x = 0, x = 1, x = 3                                        3 
           P ( x ) < 0 , x ∈ ]− 1,0[ ∪ ]1,3[ ‫ﻟﻤﺎ‬                                      2
                                                             P ( x ) < 0 , x ∈  − ∞,  ∪ ]2,+∞[ ‫ﻟﻤﺎ‬
                                                 ‫ﻟﻤﺎ‬                                  3
        P ( x ) > 0 , x ∈ ]− ∞,−1[ ∪ ]0,1[ ∪ ]3,+∞[                  P ( x ) < 0 , x ∈ ]− ∞,+∞[ ‫3( ﻟﻤﺎ‬
                                                                     P ( x ) > 0 , x ∈ ]− ∞,+∞[ ‫4( ﻟﻤﺎ‬
                  P ( x ) = ( 2 x − 3 )( x 2 + 1) (1 67
                                                                                        3 2
                                                3                       P( x ) = 0 , x =    ‫5( ﻟﻤﺎ‬
                           P ( x ) = 0 , x = ‫ﻟﻤﺎ‬                                         2
                                                2
                                                                                 3 2  3 2      
                                              3         P( x ) > 0 , x ∈  − ∞,     ∪    ,+∞  ‫ﻟﻤﺎ‬
                   P ( x ) < 0 , x ∈  − ∞,  ‫ﻟﻤﺎ‬
                                              2                                 2   2        
                                        3                                              15
                   P ( x ) > 0 , x ∈  ,+∞  ‫ﻟﻤﺎ‬                        P( x ) = 0 , x =    ‫6( ﻟﻤﺎ‬
                                        2                                              5
            P ( x ) = ( x − 1)( − x + x − 5 ) (2
                                     2                                                        ‫ﻟﻤﺎ‬
                        P ( x ) = 0 , x = 1 ‫ﻟﻤﺎ‬                                     15   15        
                                                             P( x ) > 0 , x ∈  − ∞,    ∪       ,+∞ 
                 P ( x ) < 0 , x ∈ ]1,+∞[ ‫ﻟﻤﺎ‬                                       5   5          
                 P ( x ) > 0 , x ∈ ]− ∞,1[ ‫ﻟﻤﺎ‬
                                                                          P ( x ) = 0 , x = 6 ‫7( ﻟﻤﺎ‬
                     P ( x ) = ( x − 1 )2 ( x − 2 ) (3        P ( x ) < 0 , x ∈ ]− ∞,6[ ∪ ]6,+∞[ ‫ﻟﻤﺎ‬
                   P ( x ) = 0 , x = 1, x = 2 ‫ﻟﻤﺎ‬                          P ( x ) = 0 , x = 3 ‫8( ﻟﻤﺎ‬
           P ( x ) < 0 , x ∈ ]− ∞,1[ ∪ ]1,2[ ‫ﻟﻤﺎ‬                            ]        [ ]
                                                          P ( x ) < 0 , x ∈ − ∞, 3 ∪ 3 ,+∞ ‫ﻟﻤﺎ‬   [
                     P ( x ) > 0 , x ∈ ]2,+∞[ ‫ﻟﻤﺎ‬                    P ( x ) > 0 , x ∈ ]− ∞,+∞[ ‫9( ﻟﻤﺎ‬
                                     2
          P ( x ) > 0 , x ∈  − ∞,  ∪ ]1,2[ (4                                                      66
                                     3                                                   3
                                                                         P( x ) = 0 , x =     ‫1( ﻟﻤﺎ‬
                                               2                                           2
         P ( x ) = 0 , x = 1, x = 2, x = ‫ﻟﻤﺎ‬
                                               3                                     3    
                                                                 P ( x ) > 0 , x ∈  ,+∞  ‫ﻟﻤﺎ‬
                              2                                                    2    
         P ( x ) < 0 , x ∈  ,1 ∪ ]1,+∞[ ‫ﻟﻤﺎ‬
                              3                                                         3
                                                                 P ( x ) < 0 , x ∈  − ∞,  ‫ﻟﻤﺎ‬
                                    2                                                   2
         P ( x ) > 0 , x ∈  − ∞,  ∪ ]1,2[ ‫ﻟﻤﺎ‬
                                    3                                    P ( x ) = 0 , x = 1 ‫2( ﻟﻤﺎ‬
             P ( x ) = x( x − 1)( x − 2 x − 3 ) (5
                                    2


                                                ‫ﻟﻤﺎ‬                     P ( x ) > 0 , x ∈ ]− ∞,1[ ‫ﻟﻤﺎ‬
    P ( x ) = 0 , x = −1, x = 0, x = 1, x = 3                           P ( x ) < 0 , x ∈ ]1,+∞[ ‫ﻟﻤﺎ‬
            P ( x ) < 0 , x ∈ ]− 1,0[ ∪ ]1,3[ ‫ﻟﻤﺎ‬                     P ( x ) = 0 , x = 1 , x = 2 ‫3( ﻟﻤﺎ‬
                                                    ‫ﻟﻤﺎ‬         P ( x ) < 0 , x ∈ ]− ∞,1[ ∪ ] ,2[ ‫ﻟﻤﺎ‬
                                                                                              1
 P ( x ) > 0 , x ∈ ]− ∞,−1[ ∪ ]0,1[ ∪ ]3,+∞[                            P ( x ) > 0 , x ∈ ]2,+∞[ ‫ﻟﻤﺎ‬
                                                                                               2
              P ( x ) = ( x 2 − 1)( x 2 − 2 ) (1 67          P ( x ) = 0 , x = 1 , x = 2, x =     ‫4( ﻟﻤﺎ‬
                                                                                               3
                                                ‫ﻟﻤﺎ‬                                2 
P ( x ) = 0 , x = −1, x = − 2 , x = 1, x = 2                  P ( x ) < 0 , x ∈  ,1 ∪ ]2,+∞[ ‫ﻟﻤﺎ‬
                                                                                   3 
                           ]         [ ]
      P ( x ) < 0 , x ∈ − 2 ,−1 ∪ 1,− 2 ‫ﻟﻤﺎ‬       [
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                                 ‫ﯾﻮﺟﺪ ﺣﻞ ﻣﻀﺎﻋﻒ،‬                                                                 ‫ﻟﻤﺎ‬
              ‫3‬
              ‫4‬
                 ‫4( ﻟﻤﺎ 1− = ‪ m‬ﯾﻮﺟﺪ ﺣﻠﯿﻦ 1 ،‬
                                                                                ‫]‬          ‫[‬
                                                              ‫∪ [1,1 −] ∪ 2 −,∞ − ∈ ‪P( x ) > 0 , x‬‬         ‫[∞+, 2 ]‬
        ‫ﻟﻤﺎ 1− ≠ ‪ m‬اﻟﻤﻌﺎدﻟﺔ ﺗﺼﺒﺢ ﻣﻦ اﻟﺪرﺟﺔ‬                                       ‫2( ) 4 + ‪P ( x ) = ( x − 1)( x‬‬
                                                                                               ‫2‬       ‫2‬

           ‫اﻟﺜﺎﻟﺜﺔ ﺗﻘﺒﻞ ﺛﻘﺒﻞ ﺛﻼث ﺣﻠﻮل ﻣﺘﻤﺎﯾﺰة.‬
                                                                                ‫ﻟﻤﺎ 1 = ‪P ( x ) = 0 , x = −1, x‬‬
                                        ‫07 اﻟﺸﻜﻞ اﻷول:‬
              ‫4 = ‪f ( x ) = 0 , x = −3, x = 1, x‬‬
                                                                                    ‫ﻟﻤﺎ [1;1 −] ∈ ‪P ( x ) < 0 , x‬‬
             ‫[4,1] ∪ [3−,∞ −] ∈ ‪f ( x ) < 0 , x‬‬
            ‫[∞+,4] ∪ [1,3 −] ∈ ‪f ( x ) > 0 , x‬‬                        ‫ﻟﻤﺎ [∞+,1] ∪ [1−,∞ −] ∈ ‪P ( x ) > 0 , x‬‬
                                          ‫اﻟﺸﻜﻞ اﻟﺜﺎﻧﻲ:‬
     ‫4 = ‪f ( x ) = 0 , x = −2, x = −1, x = 3, x‬‬
                                                                              ‫3( ) 4 + 2 ‪P ( x ) = ( x 2 − 2 )( 3x‬‬
                   ‫[4,3] ∪ [1−,2 −] ∈ ‪f ( x ) < 0 , x‬‬
‫[∞+,4] ∪ [3,1 −] ∪ [2−,∞ −] ∈ ‪f ( x ) > 0 , x‬‬
                                                                ‫= ‪P( x ) = 0 , x = − 2 , x‬‬                   ‫ﻟﻤﺎ 2‬
                               ‫1‪‬‬    ‫‪‬‬
                ‫‪S = ]− ∞,−3] ∪  ,+∞ ‬‬
                                                         ‫17‬                                        ‫]‬
                                                                              ‫ﻟﻤﺎ 2 ; 2 − ∈ ‪P ( x ) < 0 , x‬‬ ‫[‬
                               ‫2‪‬‬    ‫‪‬‬
                                                ‫1(‬
                                                                                     ‫]‬
                                                                 ‫−,∞ − ∈ ‪P ( x ) > 0 , x‬‬       ‫ﻟﻤﺎ [∞+, 2 ] ∪ [ 2‬
                               ‫‪‬‬    ‫‪1‬‬
                          ‫‪S = − 2, ‬‬           ‫2(‬                                                               ‫86‬
                               ‫‪‬‬    ‫‪3‬‬                                                        ‫‪‬‬    ‫‪1‬‬
                                                                         ‫) 6 + 2 ‪P ( x ) = ( x − 3 ) x − ( 2 x‬‬
                               ‫‪‬‬    ‫‪5‬‬                                                        ‫‪‬‬    ‫‪2‬‬
                          ‫‪S =  − 3, ‬‬          ‫3(‬                                                          ‫1‬
                               ‫‪‬‬    ‫‪2‬‬                                            ‫ﻟﻤﺎ = ‪P ( x ) = 0 , x = 3, x‬‬
                      ‫‪‬‬    ‫‪5‬‬                                                                               ‫2‬
                 ‫[∞+,2] ∪ ‪S =  − ∞, ‬‬          ‫4(‬                                                     ‫‪1 ‬‬
                      ‫‪‬‬    ‫‪3‬‬                                                        ‫ﻟﻤﺎ ‪P ( x ) < 0 , x ∈  ;3‬‬
                                 ‫‪S =ℜ‬‬           ‫5(‬                                                     ‫‪2 ‬‬
                                 ‫‪S =φ‬‬                                                       ‫‪‬‬     ‫‪1‬‬
                                                ‫6(‬                   ‫ﻟﻤﺎ [∞+,3] ∪ ‪P ( x ) > 0 , x ∈  − ∞, ‬‬
                                 ‫‪S =ℜ‬‬           ‫7(‬                                          ‫‪‬‬     ‫‪2‬‬
                                 ‫‪S =φ‬‬           ‫8(‬                               ‫2‬                               ‫96‬
                                                                                    ‫1( ﻟﻤﺎ 1 = ‪ m‬ﯾﻮﺟﺪ ﺣﻞ وﺣﯿﺪ‬
                                 ‫‪S =φ‬‬           ‫9(‬                               ‫3‬
                                                                                ‫ﻟﻤﺎ 1 ≠ ‪ m‬ﯾﻮﺟﺪ ﺣﻠﯿﻦ ﻣﺨﺘﻠﻔﯿﻦ‬
                                        ‫27‬
                        ‫1( [1,∞ −] = ‪S‬‬                                         ‫3‬
                                                                           ‫2( ( ﻟﻤﺎ = ‪ m‬ﯾﻮﺟﺪ ﺣﻞ وﺣﯿﺪ −‬
                                                                                                      ‫1‬
                         ‫2( [∞−,1[ = ‪S‬‬                                         ‫2‬                      ‫2‬
                ‫3( [∞+,2] ∪ [1,1 −] = ‪S‬‬                                      ‫≠ ‪ m‬ﯾﻮﺟﺪ ﺣﻠﯿﻦ ﻣﺨﺘﻠﻔﯿﻦ .‬
                                                                                                       ‫1‬
                                                                                                           ‫ﻟﻤﺎ‬
                                     ‫4(‬                                                                ‫2‬
         ‫]‬           ‫] [‬
     ‫∞+, 3 ∪ 2 , 2 − ∪ 3 −,∞ − = ‪S‬‬  ‫] [‬              ‫[‬                          ‫3( ﻟﻤﺎ 0 = ‪ m‬ﯾﻮﺟﺪ ﺣﻞ وﺣﯿﺪ 2.‬
                                                                                                ‫ﻟﻤﺎ 0 ≠ ‪. m‬و‬
                                     ‫5(‬

                                            ‫37‬                     ‫‪‬‬      ‫82 + 5 − ‪− 5 − 28  ‬‬             ‫‪‬‬
                                   ‫‪1 ‬‬                        ‫,∞ − ‪m ∈ ‬‬            ‫‪∪‬‬                ‫‪,+∞ ‬‬
                          ‫1( ‪Df = ℜ −  ,2‬‬
                                   ‫‪2 ‬‬                            ‫‪‬‬           ‫3‬     ‫‪ ‬‬        ‫3‬           ‫‪‬‬
                                     ‫‪1 ‬‬                                                         ‫ﻻ ﯾﻮﺟﺪ ﺣﻠﻮل‬
                                 ‫2( ‪S =  ‬‬                                       ‫‪ − 5 − 28 − 5 + 28 ‬‬
                                     ‫‪5‬‬                                     ‫‪m∈‬‬                ‫,‬             ‫‪‬‬
                                                         ‫47‬                       ‫‪‬‬      ‫3‬             ‫3‬      ‫‪‬‬
                      ‫‪1 − 7 1 + 7 ‬‬                                                      ‫ﯾﻮﺟﺪ ﺣﻠﯿﻦ ﻣﺘﻤﺎﯾﺰﯾﻦ.‬
                  ‫‪S=‬‬         ‫,‬      ‫1( ‪‬‬
                      ‫2 ‪‬‬         ‫‪2 ‬‬                                        ‫82 + 5 −‬               ‫82 − 5 −‬
                                                                       ‫=‪m‬‬                 ‫= ‪ m‬أو‬
                      ‫{‬
                 ‫2( 2 + 1 − ,2 − 1 − = ‪S‬‬    ‫}‬                                     ‫3‬                      ‫3‬

‫01‬
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                                            ‫′′ ‪x ′ + 4x‬‬                                    ‫‪‬‬    ‫‪1‬‬
                             ‫≤ ′ ‪ x‬ﻣﻌﻨﺎه:‬               ‫ﻟﺪﯾﻨﺎ:‬                         ‫3( ‪S = − 2, ‬‬
                                                 ‫5‬                                         ‫‪‬‬    ‫‪6‬‬
                                     ‫′′ ‪ x ≤ x‬ﺑﻌﺪ اﻟﺘﺒﺴﯿﻂ.‬                                    ‫4( ‪S = φ‬‬
                          ‫‪x‬‬ ‫′′‪′ + 4 x‬‬                                                         ‫5( ‪S = φ‬‬
                        ‫.‬             ‫و ﻧﻔﺲ اﻟﺸﺊ ﻣﻊ ′′ ‪≤ x‬‬
                               ‫5‬
                                                                                                          ‫57‬
                                                                 ‫28‬                 ‫‪ 2 + 10 ‬‬
                      ‫+‪a‬‬
                           ‫1‬
                             ‫3=‬                                                 ‫‪S=‬‬           ‫‪‬‬      ‫1(‬
                           ‫‪a‬‬                                                        ‫‪ 3 ‬‬
                      ‫0 = 1 + ‪a 2 − 3a‬‬                                    ‫‪ 30 − 6 30 + 6 ‬‬
                                                                        ‫‪S=‬‬         ‫,‬         ‫‪‬‬      ‫2(‬
                           ‫5 −3‬          ‫5 +3‬                             ‫‪‬‬    ‫42‬        ‫‪24 ‬‬
                      ‫= ′‪a‬‬       ‫= ′′‪, a‬‬
                             ‫2‬             ‫2‬                                            ‫‪1 9 ‬‬
                                                                                   ‫‪S= , ‬‬           ‫3(‬
                                    ‫5 +3‬                                                ‫‪2 2‬‬
                                ‫=‪a‬‬          ‫و ﻣﻨﮫ:‬
                                       ‫2‬                                              ‫}8,5{ = ‪S‬‬      ‫4(‬
                                                                                 ‫}945 { = ‪S‬‬
                                                                                      ‫,791‬           ‫5(‬
                                                                 ‫38‬
                                    ‫0 = ) ‪x 2 − 5(5 − x‬‬                                               ‫67‬
                                                                                        ‫7 ‪‬‬     ‫‪‬‬
                                                                         ‫1( ‪S = ]= ∞,−3[ ∪  − ,+∞ ‬‬
                                                                                        ‫3 ‪‬‬     ‫‪‬‬
                                                                                       ‫2( [∞+,1[ = ‪S‬‬
                                                                                           ‫3( }2 −{ = ‪S‬‬

                                                                                     ‫‪ − 3 + 21 ‬‬    ‫77‬
                                                           ‫58‬                   ‫‪S=‬‬             ‫‪‬‬ ‫1(‬
                            ‫0 = 3 − ‪2 x + mx‬‬
                               ‫2‬
                                                    ‫1(‬                               ‫‪‬‬      ‫2‬   ‫‪‬‬
                           ‫42 + ‪∆ = m‬‬                                            ‫{‬                   ‫}‬
                                     ‫2‬
                                                                            ‫2( 8 + 2−, 8 − 2 − = ‪S‬‬
           ‫اﻟﻤﻨﺤﻨﻲ )‪ (h‬و اﻟﻤﺴﺘﻘﯿﻢ )‪ (d‬ﯾﺘﻘﺎﻃﻌﺎن ﻓﻲ‬
                             ‫ﻧﻘﻄﺘﯿﻦ ﺣﯿﺚ ∗‪x ∈ ℜ‬‬                                                                 ‫87‬
                                                                                            ‫1( }4,3{ = ‪S‬‬
                                                 ‫2(‬
   ‫42 + 2 ‪ − m − m 2 + 24 − m − m‬‬               ‫‪‬‬                                          ‫2( }9,4{ = ‪S‬‬
‫‪M ′‬‬
   ‫‪‬‬
                       ‫,‬                      ‫‪+ m‬‬                                            ‫3( }4{ = ‪S‬‬
            ‫4‬                     ‫2‬              ‫‪‬‬
   ‫‪‬‬                                             ‫‪‬‬                                             ‫‪ 1‬‬
                                                                                           ‫4( ‪S = 3, ‬‬
    ‫42 + ‪ − m + m + 24 − m + m‬‬
                ‫2‬                    ‫2‬            ‫‪‬‬                                            ‫‪ 2‬‬
‫‪M ′′‬‬                   ‫,‬                     ‫‪+ m‬‬
    ‫‪‬‬       ‫4‬                      ‫2‬              ‫‪‬‬
    ‫‪‬‬                                             ‫‪‬‬
                                                                      ‫97 2( ﺑﻌﺪ اﻟﻨﺸﺮ و اﻟﺘﺒﺴﯿﻂ ﻧﺠﺪ أن اﻟﻤﻌﺎدﻟﺘﯿﻦ‬
 ‫‪−m m‬‬                                                                                           ‫ﻣﺘﻜﺎﻓﺌﺘﯿﻦ.‬
‫‪I‬‬      ‫‪, ‬‬
 ‫‪ 4 2‬‬                                                                                             ‫3( }4{ = ‪S‬‬
           ‫ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ‪ I‬ھﻲ اﻟﻤﺴﺘﻘﯿﻢ اﻟﺬي ﻣﻌﺎدﻟﺘﮫ:‬                                                  ‫‪1 ‬‬
                                            ‫‪Y=-2x‬‬                                              ‫4( ‪S =  ,3‬‬
                                                                                                     ‫‪2 ‬‬

             ‫68 ﻧﻔﺮض أن ﻃﻮل ﺿﻠﻊ اﻟﻤﺮﺑﻊ ‪ EBFI‬ھﻮ ‪x‬‬                                               ‫08 1( } { = ‪S‬‬
                                                                                                     ‫1‬
                                                                                            ‫2( [1,∞ −] = ‪S‬‬
                   ‫ﻧﺤﻞ اﻟﻤﻌﺎدﻟﺔ: = ) ‪x 2 + (1 − x‬‬
                                 ‫2‬ ‫2‬
                                   ‫3‬
                                                                                                    ‫18 ﻧﻔﺮض أن:‬
                                                            ‫78‬                               ‫′′ ‪x ′ + 4 x‬‬
          ‫‪S( x ) = ( 3 − x ) x + ( 5 − x ) x‬‬
                                               ‫1( ﻟﺪﯾﻨﺎ،‬                                ‫≤ ′‪x‬‬              ‫′′ ‪≤ x‬‬
                 ‫. ‪= −2 x 2 + 8 x‬‬                                                                 ‫5‬
            ‫ﺗﻜﻮن )‪ S(x‬أﻋﻈﻤﯿﺔ ﻟﻤﺎ ﺗﻜﻮن 2 = ‪x‬‬

    ‫11‬
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                                                                    ‫51‬
                                                            ‫= ‪− 2x 2 + 8x‬‬
                                                                     ‫2( ﻧﻘﻮم ﺑﺤﻞ اﻟﻤﻌﺎدﻟﺔ: 2‬
                                                              ‫5‬      ‫3‬
                                                            ‫=‪x= , x‬‬
                                                              ‫2‬      ‫2‬
                                                    ‫3(‬                                                                ‫88‬
                                                                                                ‫3‬    ‫3‬
       ‫‪x -‬‬          ‫1-‬       ‫0‬       ‫1‬                                        ‫1( − 2) − ‪P ( x ) = 2( x‬‬
                                               ‫‪+‬‬                                               ‫2‬    ‫2‬
      ‫)‪f'(x‬‬           ‫0‬       ‫0‬        ‫0‬                                                 ‫2( 2 ‪P ( X ) = 2X‬‬
              ‫‪+‬‬                              ‫‪+‬‬                                                        ‫3(‬
       ‫)‪f(x‬‬                                                    ‫‪x -‬‬                         ‫3‬
                                                                                            ‫2‬                ‫‪+‬‬
                     ‫2‬                 ‫2‬                     ‫)‪f'(x‬‬                          ‫0‬
                                                                          ‫‪-‬‬                                     ‫‪+‬‬
                       ‫3( )‪ f(x‬ﻣﻮﺟﺒﺔ ﺗﻤﺎﻣﺎ ﻋﻠﻰ ‪.ℜ‬‬             ‫)‪f(x‬‬
            ‫4( )‪ h(x)=f(x‬ﻣﻦ أﺟﻞ ﻛﻞ ﻋﺪد ﺣﻘﯿﻘﻲ ‪.x‬‬                                                 ‫3‬
                                                                                           ‫−‬
                              ‫5( ) ‪h ( x ) = f ( x‬‬                                              ‫2‬

                                                                                   ‫3‬
                                                                               ‫−‬     ‫أﺻﻐﺮ ﻗﯿﻤﺔ ﻟـ )‪ P(x‬ھﻲ:‬
                                                                                   ‫2‬
                                                                                          ‫3‬
                                                                                        ‫32 ≤ ) ‪− ≤ P ( x‬‬
                                                                                          ‫2‬
                                                                                                ‫‪1 ‬‬
                                                                                           ‫5( ‪S =  ,3‬‬
                                                                                                ‫‪2 ‬‬
                                                                          ‫‪y‬‬
                                                                                                        ‫6(‬
                          ‫‪y‬‬
                          ‫6‬                                               ‫4‬

                                                                          ‫3‬
     ‫)‪(Cg‬‬                 ‫5‬
                                           ‫)‪(Cf) (Ch‬‬                      ‫2‬

                          ‫4‬                                               ‫1‬


                                                                          ‫0‬        ‫1‬    ‫2‬           ‫3‬   ‫4‬    ‫‪x‬‬
                          ‫3‬
                                                                      ‫1-‬

                          ‫2‬                                           ‫2-‬

                                                            ‫ﻧﻼﺣﻆ أن )‪ (γ‬ﯾﻜﻮن أﺳﻔﻞ اﻟﻤﻨﺼﻒ اﻷول ﻟﻤﺎ‬
                                                                                           ‫‪1 ‬‬
                                                                                     ‫‪. x ∈  ,3‬‬
                                                                                           ‫‪2 ‬‬

                                                                                                                      ‫98‬
                                                                  ‫1( ﻣﻦ أﺟﻞ ﻧﻞ ﻋﺪد ﺣﻘﯿﻘﻲ ‪ x‬ﻣﻦ ‪:ℜ‬‬
                                                                  ‫)‪ g(-x)=g(x‬و ﻣﻨﮫ ‪ g‬زوﺟﯿﺔ.‬
                                                             ‫)‪ (Cg‬ﯾﻨﻄﺒﻖ ﻋﻠﻰ )‪ (Cf‬ﻟﻤﺎ +‪. x∈ℜ‬‬
                                                                                             ‫2(‬
                                                             ‫‪x -‬‬                      ‫1‬                ‫‪+‬‬
                                                           ‫)‪f'(x‬‬                       ‫0‬
                                                                     ‫‪-‬‬                                     ‫‪+‬‬
‫21‬                                                         ‫)‪f(x‬‬
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