Question

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

Answer 1

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

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

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

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

$=>\frac{2}{x(x-2)(x-1)}<0$

$=>xϵ(-\infty ,0)\cup (1,2)$