a: \(A=\dfrac{2x^3-2x}{-\left(x-2\right)\left(x+2\right)+x\left|x+2\right|}\)
TH1: x>-2
\(A=\dfrac{2x^3-2x}{-\left(x-2\right)\left(x+2\right)+x\left(x+2\right)}=\dfrac{2x\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(-x-2+x\right)}\)
\(=\dfrac{2x\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\cdot\left(-2\right)}=\dfrac{-x\left(x-1\right)\left(x+1\right)}{x-2}\)
\(=\dfrac{-x^3+x}{x-2}\)
TH2: x<-2
\(A=\dfrac{2x^3-2x}{-\left(x-2\right)\left(x+2\right)-x\left(x+2\right)}\)
\(=\dfrac{2x\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(-x+2-x\right)}=\dfrac{2x\left(x-1\right)\left(x+1\right)}{-2\left(x-1\right)\left(x+2\right)}\)
\(=\dfrac{-x^2-x}{x+2}\)
b: TH1: x>-2
Để A nguyên thì \(-x^3+x⋮x-2\)
=>\(x^3-x⋮x-2\)
=>\(x^3-2x^2+2x^2-4x+3x-6+6⋮x-2\)
=>\(x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{3;1;4;0;5;-1;8\right\}\)
TH2: x<-2
Để A nguyên thì -x^2-x chia hêt cho x+2
=>x^2+x chia hết cho x+2
=>x^2+2x-x-2+2 chia hêt cho x+2
=>\(x+2\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{-4;-3\right\}\)
