a) đkxđ : \(x\ne0,x\ne2,x\ne-2\)
\(A=\dfrac{2x^2+4x}{x^3-4x}+\dfrac{x^2-4}{x^2+2x}+\dfrac{2}{2-x}\)
\(A=\dfrac{2x^2+4x}{x\left(x+2\right)\left(x-2\right)}+\dfrac{x^2-4}{x\left(x+2\right)}-\dfrac{2}{x-2}\)
\(=\dfrac{2x\left(x+2\right)+\left(x-2\right)\left(x+2\right)\left(x-2\right)-2x\left(x+2\right)}{x\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{\left(x+2\right)\left(2x+\left(x-2\right)^2-2x\right)}{x\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}\)
\(=\dfrac{x-2}{x}\)
b)đkxđ : \(x\ne0,x\ne2,x\ne-2\)
với x = 4 ta có :
\(A=\dfrac{4-2}{4}=\dfrac{1}{2}\)
c)đkxđ : \(x\ne0,x\ne2,x\ne-2\)
\(A=\dfrac{x-2}{x}=1-\dfrac{2}{x}\)
\(A\in Z\Leftrightarrow\dfrac{2}{x}\in Z\Leftrightarrow x\inƯ\left(2\right)\)
\(\left[{}\begin{matrix}x=2\left(ktmdk\right)\\x=-2\left(ktmdk\right)\\x=1\left(tmdk\right)\\x=-1\left(tmdk\right)\end{matrix}\right.\)
vậy A nguyên khi x=1 hoặc x=-1