Помогите решить подробно (x/(x-1))^2+(x/(x+1))^2=45/16

$\frac{x^2}{(x-1)^2}+\frac{x^2}{(x+1)^2}=\frac{45}{16}\\\frac{x^2(x+1)^2+x^2(x-1)^2}{(x-1)^2(x+1)^2}=\frac{45}{16}\\\frac{x^2((x+1)^2+(x-1)^2)}{(x-1)^2(x+1)^2}=\frac{45}{16}\\\frac{x^2(2x^2+2)}{x^4-2x^2+1}=\frac{45}{16}\\16x^2(2x^2+2)=45(x^4-2x^2+1)\\32x^4+32x^2=45x^4-90x^2+45\\13x^4-122x^2+45=0\\x^2=t\\13t^2-122t+45=0\\D=14884-4*13*45=12544\\t_{_1}=\frac{5}{13}\\t_{_2}=9\\\\1)x^2=t\\x^2=\frac{5}{13}\\x=б\sqrt{\frac{5}{13}}\\2)x^2=t\\x^2=9\\x=б3$