a: |x-2|=1
=>x-2=1 hoặc x-2=-1
=>x=1(loại) hoặc x=3(nhận)
Khi x=3 thì \(P=\dfrac{3+2}{3-1}=\dfrac{5}{2}\)
b: \(Q=\dfrac{x^2-1+2x+1}{x\left(x+1\right)}=\dfrac{x\left(x+2\right)}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)
c: \(M=P:Q=\dfrac{x+2}{x-1}:\dfrac{x+2}{x+1}=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}\)
Để M nguyên thì \(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;-3\right\}\)