Question

\(\dfrac{x+2}{x-2}\)-\(\dfrac{x-2}{x+2}\)=\(\dfrac{4x^2}{x^2-4}\)

Answer 1

\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2-4x^2}{x^2-4}=0\) \(\left(dk:x\ne\pm2\right)\)

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

\(\Leftrightarrow-4x^2+8x=0\)

\(\Leftrightarrow-4x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=2\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)