f) \(\left(1-\dfrac{x-1}{x+1}\right)\left(x+2\right)=\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\left(x\ne\pm1\right).\)
\(\Leftrightarrow\dfrac{x+1-x+1}{x+1}\left(x+2\right)-\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=0.\)
\(\Leftrightarrow\dfrac{2\left(x+2\right)}{x+1}-\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}=0.\)
\(\Leftrightarrow\dfrac{2x+4-x+1}{x+1}-\dfrac{x+1}{x-1}=0.\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+5\right)-\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}=0.\)
\(\Rightarrow x^2+5x-x-5-\left(x^2+2x+1\right)=0.\)
\(\Leftrightarrow x^2+4x-5-x^2-2x-1=0.\)
\(\Leftrightarrow2x-6=0.\)
\(\Leftrightarrow2x=6.\)
\(\Leftrightarrow x=3\left(TMĐK\right).\)
Vậy \(S=\left\{3\right\}\) là nghiệm của phương trình.
. Ta có:
