a: ĐKXĐ: \(x\notin\left\{1;-1;-2\right\}\)
b: \(N=\left(\dfrac{1}{x+1}+\dfrac{1}{x-1}+\dfrac{x^2}{x^2-1}\right)\cdot\dfrac{x-1}{x+2}\)
\(=\left(\dfrac{1}{x+1}+\dfrac{1}{x-1}+\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x-1}{x+2}\)
\(=\dfrac{x-1+x+1+x^2}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{x-1}{x+2}\)
\(=\dfrac{x^2+2x}{\left(x+2\right)\left(x+1\right)}=\dfrac{x}{x+2}\)
c: |x|=2
=>x=2(nhận) hoặc x=-2(loại)
Thay x=2 vào N, ta được:
\(N=\dfrac{2}{2+2}=\dfrac{2}{4}=\dfrac{1}{2}\)