Решите уравнение............

$\frac{x^2}{x^2-1}= \frac{4x+5}{x^2-1} ==\ \textgreater \ x \neq 1; x \neq-1 \\ \frac{x^2}{(x-1)(x+1)} = \frac{4x+5}{(x-1)(x+1)} \\ x^2=4x+5 \\ x^2-4x-5=0 \\ D=36 \\ x1=5 \\ x2 \neq 1$
Ответ: x=5

$\frac{5}{x-3} - \frac{8}{x} =3 ==\ \textgreater \ x \neq 3; x \neq 0 \\ \frac{5x-8(x-3)-3x(x-3)}{x(x-3)} =0 \\ \frac{6x+24-3x^2}{x(x-3)} =0 \\ 6x+24-3x^2=0 \\ -3x^2+6x+24=0(:-3) \\ x^2-2x-8=0 \\ D=36 \\ x1=4 \\ x2=-2$