a, ĐK: \(x\ne\pm1\)
\(M=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\left(\dfrac{1}{x+1}-\dfrac{x}{1-x}+\dfrac{2}{x^2-1}\right)\)
\(=\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\left[\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}+\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right]\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}.\dfrac{x-1}{x+1}\)
\(=\dfrac{4x}{\left(x+1\right)^2}\)
\(M=\dfrac{1}{2}\Leftrightarrow\dfrac{4x}{\left(x+1\right)^2}=\dfrac{1}{2}\Leftrightarrow x^2-6x+1=0\Leftrightarrow x=3\pm2\sqrt{2}\)
b, \(2\left|x-3\right|=8\Leftrightarrow\left|x-3\right|=4\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\left(l\right)\end{matrix}\right.\)
\(x=7\Rightarrow M=\dfrac{7}{16}\)
a: Ta có: \(M=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\left(\dfrac{1}{x+1}-\dfrac{x}{1-x}+\dfrac{2}{x^2-1}\right)\)
\(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)
\(=\dfrac{4x}{\left(x+1\right)^2}\)
Để \(M=\dfrac{1}{2}\) thì \(\left(x+1\right)^2=8x\)
\(\Leftrightarrow x^2+2x+1-8x=0\)
\(\Leftrightarrow x^2+6x+9=8\)
\(\Leftrightarrow\left(x+3\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}-3\\x=-2\sqrt{2}-3\end{matrix}\right.\)
b: Ta có: 2|x-3|=8
\(\Leftrightarrow\left|x-3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Thay x=7 vào M, ta được:
\(M=\dfrac{4\cdot7}{\left(7+1\right)^2}=\dfrac{28}{64}=\dfrac{7}{16}\)
