Biết rằng đồ thị của hàm số

C

- Hướng dẫn giải

Đáp án C

Ta có: $T=\frac{1}{\left(x_1-1\right)\left(x_1-3\right)}+\frac{1}{\left(x_2-1\right)\left(x_2-3\right)}+\frac{1}{\left(x_3-1\right)\left(x_3-3\right)}$

$T=\frac{1}{\left(x_1-1\right)\left(x_1-3\right)}+\frac{1}{\left(x_2-1\right)\left(x_2-3\right)}+\frac{1}{\left(x_3-1\right)\left(x_3-3\right)}$ vì $\frac{1}{\left(x-1\right)\left(x-3\right)}=\frac{1}{x-3}-\frac{1}{x-1}.$

Vì $x_1,x_2,x_3$ là 3 nghiệm của phương trình $P\left(x\right)=0\Rightarrow P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right).$ 

Suy ra $P\text{'}\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)+\left(x-x_2\right)\left(x-x_3\right)+\left(x-x_3\right)\left(x-x_1\right)$ 

$\Rightarrow \frac{P\text{'}\left(x\right)}{P\left(x\right)}=\frac{\left(x-x_1\right)\left(x-x_2\right)+\left(x-x_3\right)+\left(x-x_3\right)\left(x-x_1\right)}{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}=\frac{1}{x-x_1}+\frac{1}{x-x_2}+\frac{1}{x-x_3}\left(*\right).$ 

Thay $x=1,x=3$ vào biểu thức (*), ta được $T=\frac{1}{2}\left[\frac{P\text{'}\left(x\right)}{P\left(1\right)}-\frac{P\text{'}\left(3\right)}{P\left(3\right)}\right].$