\(\frac{x+2}{x\left(x+1\right)}>1\Rightarrow\frac{x-2-x\left(x+1\right)}{x\left(x+1\right)}>0\Rightarrow\frac{x-3-x^2-x}{x\left(x+1\right)}>0\Rightarrow\frac{-x^2-3}{x\left(x+1\right)}>0\)
Lập bảng xét dấu:
x | \(-\infty\) -1 0 \(+\infty\) |
-x2 - 3 | + + + |
x | - - 0 + |
x + 1 | - 0 + + |
Vế trái | + // - // + |
Vậy S = (-\(\infty\) ; -1) \(\cup\) (0 ; +\(\infty\))