ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
Ta có: \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow5x=-25\)
hay x=-5(ktm)
Vậy: Tập nghiệm \(S=\varnothing\)
