Question

$(\frac{x+1}{x-1} )^{2} +(\frac{x}{x+5} )^{2} =\frac{2x^{2}+2x}{x^{2}+4x-5}$

Answer 1

ОДЗ:х≠1;х≠-5

$(\frac{x+1}{x-1} )^{2} +(\frac{x}{x+5} )^{2} =\frac{2x^{2}+2x}{x^{2}+4x-5}=2*\frac{x+1}{x-1} *\frac{x}{x+5}$

$(\frac{x+1}{x-1}-\frac{x}{x+5} )^{2} =0\\ \\ \frac{x+1}{x-1}-\frac{x}{x+5} =0\\ \\ \frac{(x+1)(x+5)-x(x-1)}{(x-1)(x+5)} =0\\ \\x^2+6x+5-x^2+x=0 \\ \\ 7x=-5\\ \\x=- \frac{5}{7}$