a: \(M=\left(\dfrac{x}{x+2}+\dfrac{x^3-8}{x^3+8}\cdot\dfrac{x^2-2x+4}{4-x^2}\right):\dfrac{2}{x+2}\)
\(=\left(\dfrac{x}{x+2}+\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{-\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x+2}{2}\)
\(=\left(\dfrac{x}{x+2}+\dfrac{-\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{2}\)
\(=\dfrac{x\left(x+2\right)-x^2-2x-4}{\left(x+2\right)^2}\cdot\dfrac{x+2}{2}\)
\(=\dfrac{x^2+2x-x^2-2x-4}{2\left(x+2\right)}=\dfrac{-4}{2\left(x+2\right)}=\dfrac{-2}{x+2}\)
b: \(x^2+2x=0\)
=>x(x+2)=0
=>\(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào M, ta được:
\(M=\dfrac{-2}{0+2}=\dfrac{-2}{2}=-1\)
c: Để M*x nguyên thì \(-2x⋮x+2\)
=>\(-2x-4+4⋮x+2\)
=>\(4⋮x+2\)
=>\(x+2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{-1;-3;0;-4;2;-6\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{-1;-3;0;-4;-6\right\}\)
