Sửa đề: \(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{5-x}{2x^2+10x}\)
ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
Ta có: \(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{5-x}{2x^2+10x}\)
\(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x+25}{2\left(x^2-25\right)}=\dfrac{5-x}{2x\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(2\left(x+5\right)^2-x\left(x+25\right)=-\left(x-5\right)^2\)
\(\Leftrightarrow2\left(x^2+10x+25\right)-x^2-25x=-\left(x^2-10x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2-25x=-x^2+10x-25\)
\(\Leftrightarrow x^2-5x+50+x^2-10x+25=0\)
\(\Leftrightarrow2x^2-15x+75=0\)
\(\Leftrightarrow2\left(x^2-\dfrac{15}{2}x+\dfrac{75}{2}\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{15}{4}+\dfrac{225}{16}+\dfrac{375}{16}=0\)
\(\Leftrightarrow\left(x-\dfrac{15}{4}\right)^2+\dfrac{375}{16}=0\)(vô lý)
Vậy: \(S=\varnothing\)
