\(\Leftrightarrow\dfrac{x-2}{x+2}-\dfrac{x^2+2}{x\left(x+2\right)}-\dfrac{3}{x}=0\\ \dfrac{x\left(x-2\right)}{x\left(x+2\right)}-\dfrac{x^2+2}{x\left(x+2\right)}-\dfrac{3\left(x+2\right)}{x\left(x+2\right)}=0\)
\(\Leftrightarrow x^2-2x-x^2-2-3x-6=0\)
\(\Leftrightarrow-5x-8=0\)
\(x=-\dfrac{8}{5}\)
\(ĐK:x\ne0;-2\)
\(\Leftrightarrow\dfrac{x-2}{x+2}-\dfrac{x^2+2}{x\left(x+2\right)}=\dfrac{3}{x}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)-x^2-2}{x\left(x+2\right)}=\dfrac{3\left(x+2\right)}{x\left(x+2\right)}\)
\(\Leftrightarrow x\left(x-2\right)-x^2-2=3\left(x+2\right)\)
\(\Leftrightarrow x^2-2x-x^2-2=3x+6\)
\(\Leftrightarrow-5x=8\left(voli\right)\)
\(\Leftrightarrow x=-\dfrac{8}{5}\left(tm\right)\)
Vậy \(S=\left\{-\dfrac{8}{5}\right\}\)
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne-2\end{matrix}\right.\)
\(\dfrac{x-2}{x+2}-\dfrac{x^2+2}{x^2+2x}=\dfrac{3}{x}\\ \Leftrightarrow\dfrac{x\left(x-2\right)}{x\left(x+2\right)}-\dfrac{x^2+2}{x\left(x+2\right)}-\dfrac{3\left(x+2\right)}{x\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{x^2-2x-x^2-2-3x-6}{x\left(x+2\right)}=0\\ \Rightarrow-5x-8=0\\ \Leftrightarrow x=-\dfrac{8}{5}\left(tm\right)\)
\(\dfrac{x-2}{x+2}-\dfrac{x^2+2}{x^2+2x}=\dfrac{3}{x}\left(ĐKXĐ:x\ne0;x\ne-2\right)\)
\(\Leftrightarrow\dfrac{x-2}{x+2}-\dfrac{x^2+2}{x\left(x+2\right)}=\dfrac{3}{x}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{x\left(x+2\right)}-\dfrac{x^2+2}{x\left(x+2\right)}=\dfrac{3\left(x+2\right)}{x\left(x+2\right)}\)
\(\Rightarrow x^2-2x-x^2-2=3x+6\)
\(\Leftrightarrow-2x-2-3x-6=0\)
\(\Leftrightarrow-5x-8-0\)
\(\Leftrightarrow x=\dfrac{-8}{5}\) (nhận).
-Vậy \(S=\left\{\dfrac{-5}{8}\right\}\)
