ĐKXĐ:\(x\ne\pm1\)
\(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}1-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x-1\right)\left(x+1\right)}=0\\ \Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\left(ĐKXĐ:x\ne\pm1\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow2x^2+2x-x^2+x=x^2-1\)
\(\Leftrightarrow2x^2+2x-x^2+x-x^2+1=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow x=\dfrac{-1}{3}\left(nhận\right)\)
-Vậy \(S=\left\{\dfrac{-1}{3}\right\}\)
