Question

Решите неравенство, пожалуйста

Answer 1

$\frac{lgx}{x^2-3x+2}\ \textless \ 0\; ,\; \; \; ODZ:\; x\ \textgreater \ 0\; ,x\ne 1\; ,\; x\ne 2\\\\a)\; \; \left \{ {{lgx\ \textgreater \ 0} \atop {x^2-3x+2\ \textless \ 0}} \right.\; \left \{ {{x\ \textgreater \ 1} \atop {(x-1)(x-2)\ \textless \ 0}} \right. \; \left \{ {{x\ \textgreater \ 1} \atop {x\in (1,2)}} \right. \; \; \Rightarrow \; \; x\in (1,2)\\\\b)\; \; \left \{ {{lgx\ \textless \ 0} \atop {x^2-3x+2\ \textgreater \ 0}} \right. \; \left \{ {{lgx\ \textless \ 0} \atop {(x-1)(x-2)\ \textgreater \ 0}} \right. \; \left \{ {{0<x\ \textless \ 1} \atop {x\in (-\infty ,1)\cup (2,+\infty )}} \right. \; \; \Rightarrow \; \; x\in (0,1)\\\\Otvet:\; \; x\in (0 ,1)\cup (1,2)\; .$