Question

Solve: $\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}>1$

Answer 1

$\frac{(x-1)(x-2)(x-3)-(x+1)(x+2)(x+3)}{(x+1)(x+2)(x+3)}>0$

$\frac{(x-1)(x^2-5x+6)-(x+1)(x^2+5x+6)}{(x+1)(x+2)(x+3)}>0$

$\frac{{(x}^3-{5x}^2+6x-x^2+5x-6)-(x^3+{5x}^2+6x+x^2+5x+6)}{(x+1)(x+2)(x+3)}>0$

$\frac{x^3-6x^2+11x-6-(x^3+6x^2+11x+6)}{(x+1)(x+2)(x+3)}>0$

$\frac{-12x^2-12}{(x+1)(x+2)(x+3)}>0$

$\frac{x^2+1}{(x+1)(x+2)(x+3)}<0(x^2+1>0)$

$\frac{1}{(x+1)(x+2)(x+3)}<0$

$xϵ(-\infty ,-3)\cup (-2,-1)$