a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
\(P=\left(\dfrac{x+1}{x-1}-\dfrac{4x^2}{1-x^2}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x^2-2x+1}{4x^2-4}\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{4x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{\left(x-1\right)^2}{4\left(x^2-1\right)}\)
\(=\dfrac{\left(x+1\right)^2+4x^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)^2}{4\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1+4x^2-x^2+2x-1}{\left(x+1\right)^2}\cdot\dfrac{1}{4}\)
\(=\dfrac{4x^2+4x}{4\left(x+1\right)^2}=\dfrac{4x\left(x+1\right)}{4\left(x+1\right)^2}=\dfrac{x}{x+1}\)
b: Thay x=1/3 vào P, ta được:
\(P=\dfrac{1}{3}:\left(\dfrac{1}{3}+1\right)=\dfrac{1}{3}:\dfrac{4}{3}=\dfrac{1}{3}\cdot\dfrac{3}{4}=\dfrac{1}{4}\)
c: Để P là số nguyên thì \(x⋮x+1\)
=>\(x+1-1⋮x+1\)
=>\(-1⋮x+1\)
=>\(x+1\in\left\{1;-1\right\}\)
=>\(x\in\left\{0;-2\right\}\)
