a) ĐKXĐ:\(a\ne\pm2\)
Ta có: \(A=\left(\dfrac{2}{a^2-4}+\dfrac{1}{a+2}\right):\dfrac{1}{a+2}\)
\(\Leftrightarrow A=\left(\dfrac{2}{\left(a-2\right)\left(a+2\right)}+\dfrac{1}{a+2}\right):\dfrac{1}{a+2}\)
\(\Leftrightarrow A=\left(\dfrac{2}{\left(a-2\right)\left(a+2\right)}+\dfrac{a-2}{\left(a-2\right)\left(a+2\right)}\right).\left(a+2\right)\)
\(\Leftrightarrow A=\dfrac{a}{\left(a-2\right)\left(a+2\right)}.\left(a+2\right)\)
\(\Leftrightarrow A=\dfrac{a}{a-2}\)
b)\(A\in Z\Leftrightarrow\dfrac{a}{a-2}\in Z\Leftrightarrow\dfrac{a-2+2}{a-2}\in z\Leftrightarrow\dfrac{2}{a-2}\in Z\)
\(\Leftrightarrow a-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(\Leftrightarrow a\in\left\{0;1;3;4\right\}\)(đều thoả mãn ĐKXĐ)
Vậy \(\Leftrightarrow a\in\left\{0;1;3;4\right\}\)
