a) ĐKXĐ: x∉{2;-2}
b) Ta có: \(A=\dfrac{x}{x-2}+\dfrac{2-x}{x+2}+\dfrac{12-10x}{x^2-4}\)
\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{12-10x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x-x^2+4x-4+12-10x}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-4x+8}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-4}{x+2}\)
c) Để \(A=\dfrac{2}{3}\) thì \(\dfrac{-4}{x+2}=\dfrac{2}{3}\)
\(\Leftrightarrow x+2=\dfrac{-4\cdot3}{2}=-\dfrac{12}{2}=-6\)
hay x=-6-2=-8(nhận)
Vậy: Để \(A=\dfrac{2}{3}\) thì x=-8
d) Để A nguyên thì \(-4⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(-4\right)\)
\(\Leftrightarrow x+2\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow x\in\left\{-1;-3;0;-4;2;-6\right\}\)(nhận)
Vậy: Để A nguyên thì \(x\in\left\{-1;-3;0;-4;2;-6\right\}\)