Cho đường thẳng d đi qua A(1;0) có hệ số góc k

      Tìm k để d cắt $(C):y=x^{3}-3x^{2}+2$ tại 3 điểm phân biệt có hoành độ $x_{1},x_{2},x_{3}$ thoả mãn $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=11$
gọi pt $d: y=ax+b$
pt hoành độ giao điểm: $ax+b=x^3-3x^2+2$
<=> $x^3-3x^2-ax+2-b=0$
định lý viét: $\begin{cases}x_1+x_2+x_3=3 \\ x_1 x_2+x_2 x_3 + x_3 x_1=-a \end{cases}$
$x_1^2 +x_2^2+x_3^2=11$
<=>$(x_1+x_2 +x_3)^2-2(x_1 x_2+ x_2 x_3+x_3 x_1)=11$
=> $a=1$ 
vì đt d đi qua $A(1,0)$=> $b=-1$
=>dt $d : y=x-1$
tơi1 đấy => k là $tank=a$
thì làm cho dễ hìu :v –  AKIRA 24-07-16 07:16 AM
bài này không dùng Vi-et cho bậc 3 cũng đc :)) tại có 1 nghiệm x=1 rồi =)) –  Aerilate 23-07-16 11:19 PM

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