\(\frac{3}{x-5}=\frac{-4}{x+2}\) ĐK: \(x\ne5;x\ne-2\)
\(pt\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\Leftrightarrow3x+6=-4x+20\Leftrightarrow7x=14\Leftrightarrow x=2\left(TM\right)\)
Vậy \(S=\left\{2\right\}\)
ĐK: \(\left\{{}\begin{matrix}x-5\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-2\end{matrix}\right.\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\\ \Leftrightarrow3.\left(x+2\right)=-4.\left(x-5\right)\\ \Leftrightarrow3x+6=-4x+20\\ \Leftrightarrow3x+4x=20-6\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\left(Nhận\right)\\ \rightarrow S=\left\{2\right\}\)
