66 баллов Прошу решить, также включить в решение одз.

0} \atop {log_{\frac{1}{2}}(x^2-1)>0 \right. \; \left \{ {{x\in (-\infty,-1)U(1,+\infty)} \atop {log_2(x^2-1)<0}} \right. " alt="log_3log_{\frac{1}{2}}(x^2-1)<1,\; \; OOF:\; \left \{ {{x^2-1>0} \atop {log_{\frac{1}{2}}(x^2-1)>0 \right. \; \left \{ {{x\in (-\infty,-1)U(1,+\infty)} \atop {log_2(x^2-1)<0}} \right. " align="absmiddle" class="latex-formula">

$\left \{ {{x\in (-\infty,-1)U(1,+\infty)} \atop {x^2-1<1}} \right. \; \left \{ {{x\in (-\infty,-1)U(1,+\infty)} \atop {x^2-2<0}} \right. \; \left \{ {{x\in (-\infty,-1)U(1,+\infty)} \atop {x\in (-\sqrt2,+\sqrt2)}} \right. \; \to$

$x\in (-\sqrt2,-1)U(1,\sqrt2)\\\\log_3log_{\frac{1}{2}}(x^2-1)<1$

$log_3log{\frac{1}{2}}(x^2-1)<log_33,\\\\log_{\frac{1}{2}}(x^2-1)<3$

$-log_2(x^2-1)<3,$

-3,\; -3=log_22^{-3}=log_2\frac{1}{8}\\\\x^2-1>\frac{1}{8} ,\; \; x^2-\frac{9}{8}>0\\\\(x-\frac{3}{2\sqrt2})(x+\frac{3}{2\sqrt2})>0\\\\x\in (-\infty,-\frac{3}{2\sqrt2}})U(\frac{3}{2\sqrt2},+\infty)\; ,\; \; \frac{3}{2\sqrt2}\approx 1,07\\\\Otvet:\; x\in (-\sqrt2,-\frac{3}{2\sqrt2}})U(\frac{3}{2\sqrt2}},\sqrt2)" alt="log_2(x^2-1)>-3,\; -3=log_22^{-3}=log_2\frac{1}{8}\\\\x^2-1>\frac{1}{8} ,\; \; x^2-\frac{9}{8}>0\\\\(x-\frac{3}{2\sqrt2})(x+\frac{3}{2\sqrt2})>0\\\\x\in (-\infty,-\frac{3}{2\sqrt2}})U(\frac{3}{2\sqrt2},+\infty)\; ,\; \; \frac{3}{2\sqrt2}\approx 1,07\\\\Otvet:\; x\in (-\sqrt2,-\frac{3}{2\sqrt2}})U(\frac{3}{2\sqrt2}},\sqrt2)" align="absmiddle" class="latex-formula">