a.\(P=\left(\dfrac{x+3}{x^2-1}-\dfrac{3}{x+1}\right):\left(1-\dfrac{2}{x-1}\right)\)
\(P=\left(\dfrac{x+3-3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x-1-2}{x-1}\right)\)
\(P=\left(\dfrac{x+3-3x+3}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x-3}{x-1}\right)\)
\(P=\left(\dfrac{-2x+6}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x-3}{x-1}\right)\)
\(P=\dfrac{-2\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}.\dfrac{x-1}{x-3}\)
\(P=\dfrac{-2}{x+1}\)
b.Để \(P< 0\) thì \(x+1>0\) \(\Leftrightarrow x>-1\)
c.\(Q=x.P=\dfrac{-2x}{x+1}=\dfrac{-x-x+1-1}{x+1}=\dfrac{-x-1}{x+1}+\dfrac{1-x}{x+1}=-1+\dfrac{-x+1}{x+1}=-1-\dfrac{x}{x+1}+\dfrac{1}{x+1}\)
Để P nhận giá trị nguyên thì \(\left\{{}\begin{matrix}\dfrac{x}{x+1}\in Z\\\dfrac{1}{x+1}\in Z\end{matrix}\right.\)
Xét \(\dfrac{1}{x+1}\in Z\) \(\Rightarrow x+1\in U\left(1\right)=\left\{\pm1\right\}\)
x+1 =1 => x=0
Thế \(x=0\) vào\(\dfrac{x}{x+1}\) \(\Leftrightarrow\dfrac{0}{0+1}=0\) ( nhận )
x+1 = -1 => x=-2
Thế \(x=-2\) vào \(\dfrac{x}{x+1}\) \(\Leftrightarrow\dfrac{-2}{-2+1}=-\dfrac{2}{-1}=2\) ( nhận )
Vậy \(x\in\left\{0;2\right\}\) thì Q nhận giá trị nguyên
