Question

\(\dfrac{2\left(x+1\right)}{x-1}-\dfrac{2\left(x-1\right)}{x+1}=0\)

Answer 1

\(\dfrac{2\left(x+1\right)}{x-1}-\dfrac{2\left(x-1\right)}{x+1}=0\left(ĐKXĐ:x\ne\pm1\right)\)

\(\Leftrightarrow\dfrac{2\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow2\left(x^2+2x+1\right)-2\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow2x^2+4x+2-2x^2+4x-2=0\)

\(\Leftrightarrow8x=0\)

\(\Leftrightarrow x=0\left(nhận\right)\)

-Vậy \(S=\left\{0\right\}\)