a) điều kiện : \(x\ne\pm1\)
ta có : \(P=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right)\left(\dfrac{1-x^2}{2}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right)\dfrac{\left(x^2-1\right)^2}{4}\)
\(P=\dfrac{\left(x-2\right)\left(x^2-1\right)}{4}-\dfrac{\left(x+2\right)\left(x-1\right)^2}{4}\)
\(P=\dfrac{\left(x-2\right)\left(x^2-1\right)-\left(x+2\right)\left(x-1\right)^2}{4}\)
\(P=\dfrac{x^3-x-2x^2+2-x^3+2x^2-x-2x^2+4x-2}{4}\)
\(P=\dfrac{-2x^2+2x}{4}=\dfrac{-x^2+x}{2}\)
b) ta có : \(\dfrac{P-4}{5}=x\Leftrightarrow\dfrac{\dfrac{-x^2+x}{2}-4}{5}=x\)
\(\Leftrightarrow\dfrac{-x^2+x-8}{10}=x\Leftrightarrow\dfrac{-x^2+x-8-10x}{10}=0\)
\(\Leftrightarrow-x^2-x-8x-8=0\Leftrightarrow-x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+8\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(L\right)\\x=-8\left(N\right)\end{matrix}\right.\)
vậy \(x=-8\)