a: \(Q=\left(\dfrac{a^2+4a+4-a^2+4a-4+4a^2}{\left(a-2\right)\left(a+2\right)}\right):\dfrac{a\left(a-3\right)}{5a\left(2-a\right)}\)
\(=\dfrac{4a^2+8a}{\left(a-2\right)\left(a+2\right)}\cdot\dfrac{-5\left(a-2\right)}{a-3}\)
\(=\dfrac{-20a}{a-3}\)
b: Q chia hết cho 20 thì a/a-3 là số nguyên
=>\(a-3\in\left\{1;-1;3;-3\right\}\)
=>a=4 hoặc a=6