\(\dfrac{x}{x+2}+\dfrac{x}{x-2}=\dfrac{2x+12}{x^2-4}\)
ĐKXĐ : x \(\ne\) 2; x \(\ne\) -2
\(\Leftrightarrow\dfrac{x\left(x-2\right)+x\left(x+2\right)}{x^2-4}=\dfrac{2x+12}{x^2-4}\)
\(\Rightarrow x^2-2x+x^2+2x=2x+12\)
\(\Leftrightarrow2x^2-2x-12=0\)
\(\Leftrightarrow2x^2-4x+6x-12=0\)
\(\Leftrightarrow2x\left(x-2\right)+6\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x+6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-6\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\left(l\right)\end{matrix}\right.\)
Vậy S = { -3 }
