a:
ĐKXĐ: \(x\notin\left\{0;2;-1\right\}\)
\(Q=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}-\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\dfrac{1}{x^2-x+1}+\dfrac{1}{x^2-x+1}-\dfrac{2}{x+1}\right):\dfrac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\left(\dfrac{2}{x^2-x+1}-\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\)
\(=1+\dfrac{2x+2-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\dfrac{-2x^2+4x}{\left(x+1\right)\cdot x\left(x-2\right)}\)
\(=1+\dfrac{-2x\left(x-2\right)}{\left(x+1\right)\cdot x\cdot\left(x-2\right)}=1-\dfrac{2}{x+1}=\dfrac{x-1}{x+1}\)
b: \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Khi x=-1/2 thì \(Q=\dfrac{-\dfrac{1}{2}-1}{-\dfrac{1}{2}+1}=\dfrac{-3}{2}:\dfrac{1}{2}=-3\)
c: 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: \(x\in\left\{-2;-3;1\right\}\)
