Question

Решить систему неравенств
1) 2^x - 1 <= 0
2) log0.5 * x^2 - 1 >= 0

Answer 1

========== 1 ==========
$2^x-1 \leq 0\\ 2^x \leq 1\\ 2^x \leq 2^0\\ 2\ \textgreater \ 1 \Rightarrow \\ x \leq 0$

========== 2 ==========
$log_{0.5}x^2-1 \geq 0\\ ===============\\ x^2\ \textgreater \ 0\\ x\in R, x\neq0\\ ===============\\ log_{0.5}x^2 \geq 1\\ log_{0.5}x^2 \geq log_{0.5}0.5\\ 0\ \textless \ 0.5\ \textless \ 1 \Rightarrow \\ x^2 \leq 0.5\\ x^2-0.5 \leq 0\\ x^2- \frac{1}{2} \leq 0\\ (x- \frac{1}{ \sqrt{2} } )(x+ \frac{1}{ \sqrt{2} } ) \leq 0\\ x \in [- \frac{1}{ \sqrt{2} }; \frac{1}{ \sqrt{2} }]\\ ===============\\ \left \{ {{x\in R, x\neq0} \atop {x \in [- \frac{1}{ \sqrt{2} }; \frac{1}{ \sqrt{2} }]} \right.$
$x\in [- \frac{1}{ \sqrt{2} }; 0)\cup(0;\frac{1}{ \sqrt{2} }]$