a) điều kiện xát định : \(x\ne\pm1\)
ta có : \(P=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right).\left(\dfrac{1-x^2}{2}\right)^2\)
\(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right).\dfrac{\left(1-x\right)^2\left(1+x\right)^2}{4}\)
\(P=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{4}-\dfrac{x+2}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{4}\)
\(P=\dfrac{\left(x-2\right)\left(x-1\right)\left(x+1\right)}{4}-\dfrac{\left(x+2\right)\left(x-1\right)^2}{4}\)
\(P=\dfrac{\left(x-2\right)\left(x-1\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)^2}{4}\)
\(P=\dfrac{\left(x-1\right)\left(\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)\right)}{4}\)
\(P=\dfrac{\left(x-1\right)\left(x^2-x-2-\left(x^2+x-2\right)\right)}{4}\)
\(P=\dfrac{\left(x-1\right)\left(x^2-x-2-x^2-x+2\right)}{4}=\dfrac{\left(x-1\right)\left(-2x\right)}{4}\)
\(P=\dfrac{-2x^2+2x}{4}\)
b) ta có : \(P-4=5x\Leftrightarrow\dfrac{-2x^2+2x}{4}-4=5x\)
\(\Leftrightarrow\dfrac{-2x^2+2x-16}{4}=5x\Leftrightarrow-2x^2+2x-16=20x\)
\(\Leftrightarrow20x-\left(-2x^2+2x-16\right)=0\Leftrightarrow2x^2+18x+16=0\)
\(\Leftrightarrow2x^2+2x+16x+16=0\Leftrightarrow2x\left(x+1\right)+16\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x+16\right)\left(x+1\right)\Leftrightarrow\left[{}\begin{matrix}2x+16=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(tmđk\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
vậy \(x=-8\) thỏa mãng điều kiện bài toán
