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Phan Nguyeãn Phaùt Thaønh



12T3 THPT Chu Vaên An



Baøi 3: b

1 y  x3  2 x 2  x  4 3



y '  x2  4x  1



phương trình cần tìm có dạng: y  f '( x0 )( x  x0 )  y0 Ta có : f '( x0 )  tan 600   3

 k  3



 khi

2



k 3



 x0  4 x0  1  3  0

x  2  3  3 0  x  2  3  3  0



* Khi x0  2  3  3

y0 



6  3 



3  3  22 3



PTTT là :



y  3 x  2 3 3 











6  2 3  3x 







6  3 

3



3  3  22 3



3  3  6 3  22



* Khi x0  2  3  3

y0 







3 6







3  3  22 3



PTTT có là



y  3 x  2 3 3











 

3



3 6







3  3  22 3



 3x 



 2



3 6







3  3  6 3  22



http://www.thptchuvanan.net/forum/index.php/topic,48.345.html http://www.thptchuvanan.net/forum/index.php/topic,641.60.html



Phan Nguyeãn Phaùt Thaønh



12T3 THPT Chu Vaên An



 Khi k   3

 x0 2  4 x0  1  3  0



x  2  3  3 0  x  2  3  3  0



 Khi



x0  2  3  3



y0 



 6  3 



3  3  22 3



PTTT là :



y   3 x  2 3 3 







2 3x 











 6  3 



3  3  22 3



3 6







3  3  6 3  22 3



 khi x0  2  3  3

y0 



6  3 



3  3  22 3



PTTT là :



y   3 x  2 3 3 







  3x 



6  2 3 







6  3 



3  3  22 3



3  3  6 3  22 3



http://www.thptchuvanan.net/forum/index.php/topic,48.345.html http://www.thptchuvanan.net/forum/index.php/topic,641.60.html




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