Ta có :
\(\dfrac{a}{x-1}+\dfrac{bx+x}{x^2-x+1}=\dfrac{ax^2-ax+a+bx^2-bx+x^2-x}{\left(x-1\right)\left(x^2-x+1\right)}\)
= \(\dfrac{x^2\left(a+b+1\right)-x\left(a+b+1\right)+a}{\left(x-1\right)\left(x^2-x+1\right)}\)
Đồng nhất hai vế , ta có :
\(x^2\left(a+b+1\right)-x\left(a+b+1\right)+a=1\)
Suy ra :
* a + b +1 = 0 => 2 + b = 0 => b = - 2
* a = 1
Vậy,....