a: ĐKXĐ: x<>1; x<>-1; x<>1/2
\(C=\left(\dfrac{1}{x+1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x-1\right)^2}{2x-1}\)
\(=\dfrac{x-1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)^2}{2x-1}=\dfrac{x-1}{x+1}\)
b: Khi x=-3 thì \(C=\dfrac{-3-1}{-3+1}=\dfrac{-4}{-2}=2\)
c Để C nguyên thì x+1-2 chia hết cho x+1
=>\(x+1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{0;-2;-3\right\}\)