Ta có :
\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+3x^2+3x+1+5x^2+14x+9}\) \(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x^2+14x+9}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x^2+5x+9x+9}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x\left(x+1\right)+9\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x-2\right)}{\left(x+1\right)^3+\left(x+1\right)\left(5x+9\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x^2+7x+10}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x^2+2x+5x+10}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)+5\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}\)
\(=\dfrac{x-2}{x+5}\)
\(\Rightarrow a=-2;b=5\)
Vậy a + b = -2 + 5 = 3
!!!!