\(\dfrac{H\left(x\right)}{x-1}=\dfrac{ax^3+ax^2+x^2-4bx-3x+5b}{x-1}\)
\(=\dfrac{ax^3-ax^2+x^2\cdot\left(2a+1\right)-2ax-x+\left(2a-4b-2\right)x-2a+4b+2+b-2+2a}{x-1}\)
\(=ax^2+x\left(2a+1\right)+\left(2a-4b-2\right)+\dfrac{b+2a-2}{x-1}\)
\(\dfrac{H\left(x\right)}{x+2}\)
\(=\dfrac{ax^3+\left(a+1\right)x^2-\left(4b+3\right)x+5b}{x+2}\)
\(=\dfrac{ax^3+2ax^2+x^2\left(-a+1\right)+x\cdot\left(-2a+2\right)+[-x\left(-2a+2\right)-\left(4b+3\right)x]+5b}{x+2}\)
\(=ax^2+\left(-a+1\right)\cdot x+\dfrac{\left[2ax-2x-4bx-3x\right]+5b}{x+2}\)
\(=ax^2-ax+x+\dfrac{-5x+2ax-4bx-10+4a-8b+10-4a+13b}{x+2}\)
\(=ax^2-ax+x+\left(2a-4b-5\right)+\dfrac{-4a+13b+10}{x+2}\)
Theo đề, ta có hệ:
-4a+13b=-10 và b+2a=2
=>a=6/5; b=-2/5
