Question

Solve the following inequality:
$\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)}>1$

Answer 1

$=>\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)}>1$

$=>\frac{(x-2)(x-4)(x-7)-1}{(x+2)(x+4)(x+7)}>0$

$=>\frac{{(x}^2-6x+8\left)\right(x-7)-(x^2+6x+8\left)\right(x+7)}{(x+2)(x+4)(x+7)}>0$

$=>\frac{x^2-7x^2-{6x}^2+42x+8x-56-x^2-{6x}^2-8x-{7x}^2-42x-56}{(x+2)(x+4)(x+7)}>0$

$=>\frac{-26x^2-112}{(x+2)(x+4)(x+7)}>0$

$=>\frac{26x^2+112}{(x+2)(x+4)(x+7)}<0$

$=>xϵ(-\infty ,-7)\cup (-4,-2)$.