ОДЗ
\left \{ {{x \neq -5} \atop {x \neq 5}}\right.\\\ \\\
\frac{x}{x+5}+\frac{x+5}{x-5}=\frac{50}{x^{2}-25}\\\ \\\
\frac{x(x-5)+(x+5)(x+5)}{x^{2}-25}=\frac{50}{x^{2}-25}\\\
x(x-5)+(x+5)(x+5)=50\\\
x^2-5x+x^2+10x+25=50\\\
2x^2+5x-25=0\\\
D=25+200=225\\\
x_1=\frac{-5+15}{4}=2,5\ \ \ \ \ \ x_2=\frac{-5-15}{4}=-5" alt=" \left \{ {{x+5 \neq 0} \atop {x-5 \neq 0}}\right. \ <=> \left \{ {{x \neq -5} \atop {x \neq 5}}\right.\\\ \\\
\frac{x}{x+5}+\frac{x+5}{x-5}=\frac{50}{x^{2}-25}\\\ \\\
\frac{x(x-5)+(x+5)(x+5)}{x^{2}-25}=\frac{50}{x^{2}-25}\\\
x(x-5)+(x+5)(x+5)=50\\\
x^2-5x+x^2+10x+25=50\\\
2x^2+5x-25=0\\\
D=25+200=225\\\
x_1=\frac{-5+15}{4}=2,5\ \ \ \ \ \ x_2=\frac{-5-15}{4}=-5" align="absmiddle" class="latex-formula">
x=-5 не удовлетворяет ОДЗ
ответ: 2,5