a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: Để E là số nguyên thì \(x^2-1+1⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1\right\}\)
hay \(x=2\)
c: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>2x=4 hoặc 2x=-6
=>x=2 hoặc x=-3
Khi x=2 thì \(E=\dfrac{2^2}{2-1}=4\)
Khi x=-3 thì \(E=\dfrac{9}{-3-1}=\dfrac{-9}{4}\)
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