Question

Solve for x: $\sqrt{\frac{x-2}{1-2x}}>-1$

Answer 1

$\sqrt{x}>0$ for all real values of x and if $x<0,\sqrt{x}$ doesn't exist.

$\frac{x-2}{1-2x}>0\frac{x-2}{2x-1}<0\frac{(x-2)}{(x-\frac{1}{2})}<0x\in (\frac{1}{2},2)if\frac{x-2}{1-2x}>0\sqrt{\frac{x-2}{1-2x}}>0>-1$