Có: P=\(\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x^2-x}\right)\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\frac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\frac{x}{x\left(x-1\right)}+\frac{2-x^2}{x\left(x-1\right)}\right]\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\frac{x^2-1}{x\left(x-1\right)}+\frac{x}{x\left(x-1\right)}+\frac{2-x^2}{x\left(x-1\right)}\right]\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\frac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right]\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}.\left[\frac{x\left(x-1\right)}{x+1}\right]\) ĐKXĐ: \(x\ne\pm1\)
P=\(\frac{x^2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
P=\(\frac{x^2}{x-1}\)
Để \(P=\frac{-1}{2}\)thì\(\frac{x^2}{x-1}\)=\(\frac{-1}{2}\)
⇔2x2=-1(x-1)
⇔2x2=-x+1
⇔2x2+x-1=0
⇔2x2+2x-x-1=0
⇔2x(x+1)-(x+1)=0
⇔(2x-1)(x+1)=0
⇔\(\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{Thoả mãn}\right)\\x=-1\left(\text{Thoả mãn}\right)\end{matrix}\right.\)
Vạy để P=\(\frac{-1}{2}\)thì x=\(\frac{1}{2}\)hoặc x=-1
