Question

Решите неравенства:
$lg( x^{2} -x+8)\ \textgreater \ 1$
$lg2x\ \textless \ lg(x+1)$

Answer 1

$1)\; lg(x^2-x+8)\ \textgreater \ 1\; ,\\\\ODZ:\; x^2-x+8\ \textgreater \ 0\; ;D<0\to \; x\in R\\\\x^2-x+8\ \textgreater \ 10\\\\x^2-x-2\ \textgreater \ 0\\\\(x+1)(x-2)\ \textgreater \ 0\\\\+++(-1)---(2)+++\\\\x\in (-\infty,-1)U(2,+\infty)$

$2)\; lg2x\ \textless \ lg(x+1)\; ,\; \; ODZ:\; \left \{ {{2x\ \textgreater \ 0} \atop {x+1\ \textgreater \ 0}} \right. \; , \left \{ {{x\ \textgreater \ 0} \atop {x\ \textgreater \ -1}} \right. \; \to \; x\ \textgreater \ 0\\\\2x\ \textless \ x+1\\\\x\ \textless \ 1\\\\0\ \textless \ x\ \textless \ 1\\\\x\in (0,1)$