Решите 4 срочно!!!!!! даю все 98 баллов!!!! срочнО!!!!!!!!!!!

Уравнение 4)

$\displaystyle \frac{x+1}{x-1}+\frac{x+5}{x-5}=\frac{x+3}{x-3}+\frac{x+4}{x-4}\\\\\frac{(x+1)(x-5)+(x-1)(x+5)}{(x-1)(x-5)}=\frac{(x+3)(x-4)+(x-3)(x+4)}{(x-3)(x-4)}\\\\\frac{2x^2-10}{(x-1)(x-5)}=\frac{2x^2-24}{(x-3)(x-4)}\\\\\frac{x^2-5}{(x-1)(x-5)}=\frac{x^2-12}{(x-3)(x-4)}\\\\(x^2-5)(x-3)(x-4)=(x^2-12)(x-1)(x-5)$

$\displaystyle (x^2-5)(x^2-7x+12)=(x^2-12)(x^2-6x+5)\\\\x^4-7x^3+7x^2+35x-60=x^4-6x^3-7x^2+72x-60\\\\-x^3+14x^2-37x=0\\\\-x(x^2-14x+37)=0\\\\x_1=0; x^2-14x+37=0\\\\D=14^2-4*37=196-148=48=(4\sqrt{3})^2\\\\x_{2.3}=\frac{14\pm 4\sqrt{3}}{2}\\\\x_2=7+2\sqrt{3}; x_3=7-2\sqrt{3}$