a: ĐKXĐ: \(x\notin\left\{0;3;1\right\}\)
\(A=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
\(=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\right)\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{x^2-6x+9-x^2+9}{x-3}\cdot\dfrac{1}{2\left(x-1\right)}\)
\(=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}\)
\(=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\)
b: Để A=2 thì \(-\dfrac{3}{x-1}=2\)
=>\(x-1=-\dfrac{3}{2}\)
=>\(x=-\dfrac{3}{2}+1=-\dfrac{1}{2}\)(nhận)
c: Để A là số nguyên thì \(-3⋮x-1\)
=>\(x-1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{2;0;4;-2\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;4;-2\right\}\)
