sukonnikoffma Пусть: x/(x^2+x-5)=t
1/t+3t+4=0 |*t≠0
3t^2+4t+1=0
(3t+1)(t+1)=0
1)t=-1
x/(x^2+x-5)=-1|*(x^2+x-5)≠0
x=-x^2-x+5
x^2+2x-5=0
D1=1+5=6
x1= -1+sqrt(6)
x2=-1-sqrt(6)
2) t=-1/3
x/(x^2+x-5)=-1/3|*3(x^2+x-5)≠0
3x=-x^2-x+5
x^2+4x-5=0
(x+5)(x-1)=0
x1=-5
x2=1
Ответ:-1+sqrt(6)+
-1-sqrt(6)+ -5+ 1=-6