biết đồ thị hàm số f(x) = a(x^3)+b(x^2)+cx+d cắt trục hoành tại 3 điểm phân biệt

* Hướng dẫn giải

Đáp án D

Vì $x_1,x_2,x_3$ là ba nghiệm của phương trình $f\left(x\right)=0\Rightarrow f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right).$ 

Ta có $f\text{'}\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)+a\left(x-x_2\right)\left(x-x_3\right)+a\left(x-x_3\right)\left(x-x_1\right).$ 

Khi đó

$\begin{array}{l}\left\{\begin{array}{l}f\text{'}\left(x_1\right)=a\left(x_1-x_2\right)\left(x_1-x_3\right)\\ f\text{'}\left(x_2\right)=a\left(x_2-x_3\right)\left(x_2-x_1\right)\\ f\text{'}\left(x_3\right)=a\left(x_3-x_1\right)\left(x_3-x_2\right)\end{array}\right.\Rightarrow T=\frac{1}{a\left(x_1-x_2\right)\left(x_1-x_3\right)}+\frac{1}{a\left(x_2-x_3\right)\left(x_2-x_1\right)}+\frac{1}{a\left(x_3-x_1\right)\left(x_3-x_2\right)}\\ =\frac{1}{a\left(x_1-x_2\right)\left(x_1-x_3\right)}-\frac{1}{a\left(x_1-x_2\right)\left(x_2-x_3\right)}+\frac{1}{a\left(x_1-x_3\right)\left(x_2-x_3\right)}=\frac{x_2-x_3-x_1+x_3+x_1-x_2}{a\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}=0.\end{array}$