a: ĐKXĐ: \(x\notin\left\{1;0;-1\right\}\)
\(Q=\left[\dfrac{\left(x-1\right)^2}{x^2+x}+1-\dfrac{1}{x}\right]:\left(\dfrac{x^3-1}{x^2-x}-\dfrac{x^3+1}{x^2+x}\right)\)
\(=\left[\dfrac{\left(x-1\right)^2}{x\left(x+1\right)}+\dfrac{x\left(x+1\right)-\left(x+1\right)}{x\left(x+1\right)}\right]:\left(\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2-2x+1+x^2+x-x-1}{x\left(x+1\right)}:\dfrac{x^2+x+1-x^2+x-1}{x}\)
\(=\dfrac{2x^2-2x}{x\left(x+1\right)}\cdot\dfrac{x}{2x}=\dfrac{2x\left(x-1\right)}{\left(x+1\right)\cdot2x}=\dfrac{x-1}{x+1}\)
b: Để Q nguyên thì \(x-1⋮x+1\)
=>\(x+1-2⋮x+1\)
=>\(-2⋮x+1\)
=>\(x+1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{0;-2;1;-3\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{-2;-3\right\}\)
c: \(P=\dfrac{\left(x-1\right)^2}{Q}+x+1=\left(x-1\right)^2:\dfrac{x-1}{x+1}+x+1\)
\(=\left(x-1\right)\left(x+1\right)+x+1=x^2+x=\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(x+\dfrac{1}{2}=0\)
=>\(x=-\dfrac{1}{2}\)
