\(ĐKXĐ:x\ne\pm5\)
\(\frac{5}{x+5}-\frac{x-3}{5-x}=\frac{2x-40}{x^2-25}\)
\(\Leftrightarrow\frac{5}{x+5}+\frac{x-3}{x-5}-\frac{2x-40}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{5x-25+x^2+2x-15-2x+40}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\( \Leftrightarrow\left[{}\begin{matrix}x=-5\left(ktm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
ĐKXĐ: x∉{-5;5}
Ta có: \(\frac{5}{x+5}-\frac{x-3}{5-x}=\frac{2x-40}{x^2-25}\)
\(\Leftrightarrow\frac{5}{x+5}+\frac{x-3}{x-5}=\frac{2x-40}{x^2-25}\)
\(\Leftrightarrow\frac{5\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{\left(x-3\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{2x-40}{\left(x-5\right)\left(x+5\right)}\)
Suy ra: \(5\left(x-5\right)+\left(x-3\right)\left(x+5\right)=2x-40\)
\(\Leftrightarrow5x-25+x^2+2x-15=2x-40\)
\(\Leftrightarrow x^2+7x-40-2x+40=0\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-5\left(ktm\right)\end{matrix}\right.\)
Vậy: S={0}
