a: \(C=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\left(2-\dfrac{x+5}{x+3}\right)\)
\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)
\(=\dfrac{-5x-5}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}=\dfrac{-5}{x-3}\)
b: |x|=1
=>x=1(nhận) hoặc x=-1(loại)
Thay x=1 vào C, ta được:
\(C=\dfrac{-5}{1-3}=\dfrac{-5}{-2}=\dfrac{5}{2}\)
c: Để C là số nguyên thì \(x-1\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{2;0;6;-4\right\}\)