Question

СРОЧНО!!!!! Решить 2 уравнение

Answer 1

$\dfrac{5}{x^2-4}-\dfrac{8}{x^2-1}=\dfrac{2}{x^2-3x+2}-\dfrac{20}{x^2+3x+2}\\\\\\\dfrac{5}{(x-2)(x+2)}-\dfrac{8}{(x-1)(x+1)}=\dfrac{2}{(x-1)(x-2)}-\dfrac{20}{(x+1)(x+2)}\\\\\\\dfrac{5(x^2-1)-8(x^2-4)}{(x-2)(x+2)(x-1)(x+1)}=\dfrac{2(x^2+3x+2)-20(x^2-3x+2)}{(x-1)(x-2)(x+1)(x+2)}\\\\\\\dfrac{-3x^2+27}{(x-2)(x+2)(x-1)(x+1)}=\dfrac{-18x^2+29x-36}{(x-2)(x+2)(x_1)(x+1)}\\\\\\-3x^2+27=-18x^2+29x-36\\\\15x^2-29x+63=0\\\\D=29^2-4\cdot 15\cdot 63=-2939<0\; \; \; \Rightarrow \; \; \; x\in \varnothing$