Question

Solve $\frac{2^{1-x}-2^x+1}{2^x-1}\le 0$

Answer 1

$\frac{{2.2}^{-x}-2^x+1}{2^x-1}\le 0$

$Let{2}^x=y$

$=\frac{\frac{2}{y}}{y-1}-y+1\le 0$

$=\frac{2-y^2+y}{y\left(y-1\right)}\le 0$

$=\frac{-\left(y^2-y-2\right)}{y\left(y-1\right)}\ge 0$

$=\frac{y^2-y-2}{y\left(y-1\right)}\ge$

$=\frac{\left(y+1\right)\left(y-2\right)}{y\left(y-1\right)}\ge 0$

$=y\in \left(-\infty ,-1\right)\cup \left(0,1\right)\cup \left[2,\infty \right)$

$=2^x\in \left(0,1\right)or\left[2,\infty \right).$

$=x\in \left(1,2\right)or\left[4,\infty \right)$

$=x\in \left(1,2\right)\left[4,\infty \right)$