1/x+1/(x+1)>1/(x+2) ОДЗ: х≠0 х≠-1 х≠-2
((x+1)(x+2)+x(x+2)-x(x+1))/(x(x+1(x+2))>0
(x²+3x+2+x²+2x-x²-x)/(x(x+1)(x+2))>0
(x²+4x+2)/(x(x+1)x+2))>0
(x+2+√2)(x+2-√2)/(x(x+1)(x+2)>0
-2+√2≈-0,6 -2-√2≈-3,4 ⇒
-∞___-___-3,4___+___-2___-___-1___+___-0,6___-___0___+___+∞
x∈(-2-√2;-2)U(-1;-2+√2)U(0;+∞).