Question

Рациональные неравенства

Answer 1

4/(x+1)≥(4/(3x-3)+5/(3x+6)

ОДЗ: x+1≠0    x≠-1       x-1≠0   x≠1     x+2≠0     x≠-2

4/(x+1)≥4/(3*(x-1))+5/(3*(x+2))  |×3

4*3/(x+1)≥4/(x-1)+5/(x+2)

12/(x+1)-4/(x-1)-5/(x+2)≥0

(12*(x-1)*(x+2)-4*(x+1)*(x+2)-5*(x+1)*(x-1))/((x+1)*(x-1)*(x+2))≥0

(12x²+12x-24-4x²-12x-8-5x²+5)/((x+1)*(x-1)*(x+2)x+1))≥0

(3x²-27)/((x+1)*(x-1)*(x+2))≥0

3*(x²-9)/((x+1)*(x-1)*(x+2)≥0  |÷3

(x+3)*(x-3)/((x+1)*(x-1)*(x+2))≥0

-∞__-__-3__+__-2__-__-1__+__1__-__3__+__+∞

x∈[-3;-2)U(-1;1)U[3;+∞).    ⇒

Ответ: x=-3.