ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
Ta có: \(\frac{x+25}{2x^2-50}-\frac{x+5}{x^2-5x}=\frac{5-x}{2x^2+10x}\)
\(\Leftrightarrow\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}-\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=0\)
Suy ra: \(x^2+25x-2\left(x^2+10x+25\right)+x^2-10x+25=0\)
\(\Leftrightarrow2x^2+15x+25-2x^2-20x-50=0\)
\(\Leftrightarrow-5x-25=0\)
\(\Leftrightarrow-5x=25\)
hay x=-5(loại)
Vậy: \(S=\varnothing\)
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