a, \(P=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{1}{x^2-1}\right)\)ĐK : \(x\ne\pm1\)
\(=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x-1+x\left(x+1\right)+1}{\left(x+1\right)\left(x-1\right)}\right)\)
\(=\frac{4x}{x-1+x^2+x+1}=\frac{4x}{x^2+2x}=\frac{4}{x+2}\)
b, Thay x = -11 ta được : \(\frac{4}{-11+2}=-\frac{4}{9}\)
c, \(P\ge1\Leftrightarrow\frac{4}{x+2}-1\ge0\Leftrightarrow\frac{4-x-2}{x+2}\ge0\)
\(\Leftrightarrow\frac{2-x}{x+2}\ge0\Leftrightarrow\frac{x-2}{x+2}\le0\)
Vì \(x+2>x-2\Rightarrow\hept{\begin{cases}x+2\ge0\\x-2\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-2\\x\le2\end{cases}}\Leftrightarrow-2\le x\le2\)
Kết hợp với đk vậy \(-2\le x\le2;x\ne\pm1\)
