Решите неравенство: корень 2^х-2/2^х-4<1

Решите задачу:

$\sqrt{\frac{2^{x}-2}{2^{x}-4} }\ \textless \ 1\; ;\; \; \; \; ODZ:\; 2^{x}-4\ne 0\; ;\; 2^{x}\ne 2^2\; ;\; x\ne 2$

$\frac{2^{x}-2}{2^{x}-4} \geq 0\\\\+++[2]---(4)+++\\\\ \left \{ {{2^{x}\ \textgreater \ 4} \atop {2^{x} \leq 2}} \right. \left \{ {{2^{x}\ \textgreater \ 2^2} \atop {x \leq 1}} \right. \left \{ {{x\ \textgreater \ 2} \atop {x \leq 1}} \right. ;x\in (-\infty ,1]\cup (2,+\infty )$

$\frac{2^{x}-2}{2^{x}-4} -1\ \textless \ 0\\\\ \frac{2^{x}-2-2^{x}+4}{2^{x}-4} \ \textless \ 0\\\\ \frac{2}{2^{x}-4} \ \textless \ 0\\\\Tak\; kak\; \; 2\ \textgreater \ 0\; ,\; to\; \; 2^{x}-4\ \textless \ 0\\\\2^{x}\ \textless \ 2^2\\\\x\ \textless \ 2\\\\Otvet:\; \; x\in (-\infty ,1]\; .$