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Решите задачу:

$\cfrac{2x^2-1}{x-8}\ \textgreater \ 0\ \textless \ =\ \textgreater \ \cfrac{x^2-\frac{1}{2}}{x-8}\ \textgreater \ 0\ \textless \ =\ \textgreater \ \cfrac{(x-\sqrt{\frac{1}{2}})(x+\sqrt{\frac{1}{2}})}{x-8}\ \textgreater \ 0\\\\OTBET:x\in(-\frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2})(8;+\infty)$

$lg(2x+1)\ \textless \ 0\ \textless \ =\ \textgreater \ \left\{{{2x+1\ \textgreater \ 0}\atop{2x+1\ \textless \ 1}}\right\to\left\{{{2x\ \textgreater \ -1}\atop{2x\ \textless \ 0}}\right\to\left\{{{x\ \textgreater \ -\frac{1}{2}}\atop{x\ \textless \ 0}}\right\\\\OTBET:x\in(-\frac{1}{2};0)$

$log_{0,5}(2x)\ \textgreater \ 2\ \textless \ =\ \textgreater \ log_2(8x)\ \textless \ 0\to\left\{{{8x\ \textgreater \ 0}\atop{8x\ \textless \ 1}}\right\to\left\{{{x\ \textgreater \ 0}\atop{x\ \textless \ \frac{1}{8}}}\right\\\\OTBET:x\in(0;\frac{1}{8})$

$lg(4x-2)=2lg4-lg2\ \textless \ =\ \textgreater \ lg(4x-2)=lg8\to4x-2=8 \to x=2,5$