Question

Solve the following inequality:
$\frac{\sqrt{2x-1}}{x-2}<1.$

Answer 1

$\frac{\sqrt{2x-1}}{x-2}<1$ $2x-1>0$,$x>\frac{1}{2}$

squaring bothsides

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

$\frac{2x-1-x^2+4x-4}{(x-2{)}^2}<0$

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

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

$x\in (-\infty ,1)\cup (5,\infty )$

$\therefore x\in (\frac{1}{2},1)\cup (5,\infty )$