1: ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)
\(P=\left(\dfrac{x+1}{x^2-x}+\dfrac{1}{1-x}\right):\left(\dfrac{-2}{1-x^2}+\dfrac{1}{1+x}\right)\)
\(=\left(\dfrac{x+1}{x\left(x-1\right)}-\dfrac{1}{x-1}\right):\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+1}\right)\)
\(=\dfrac{x+1-x}{x\left(x-1\right)}:\dfrac{2+x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\)
\(=\dfrac{1}{x}\)
2:
ĐKXĐ chung của P và Q là \(x\notin\left\{0;1;-1\right\}\)
Đặt \(A=P\cdot Q=\dfrac{2x^2}{x-1}\cdot\dfrac{1}{x}=\dfrac{2x^2}{x\left(x-1\right)}=\dfrac{2x}{x-1}\)
Để A là số nguyên dương thì \(\left\{{}\begin{matrix}2x⋮x-1\\\dfrac{2x}{x-1}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-2+2⋮x-1\\\dfrac{x}{x-1}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2⋮x-1\\\left[{}\begin{matrix}x>1\\x< 0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1\in\left\{1;-1;2;-2\right\}\\\left[{}\begin{matrix}x>1\\x< 0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{2;0;3;-1\right\}\\\left[{}\begin{matrix}x>1\\x< 0\end{matrix}\right.\end{matrix}\right.\)
=>\(x\in\left\{2;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;3\right\}\)
