Question

Помогите решить систему неравенств: x^2-5x+6>=0 1/x < 1

Answer 1

$\left \{ {{x^2-5x+6 \geq o} \atop {\frac{1}{x}\ \textless \ 1}} \right. \; \left \{ {{(x-2)(x-3) \geq 0} \atop { \frac{1-x}{x} \ \textless \ 0}} \right. \; \left \{ {{(x-2)(x-3) \geq 0} \atop {\frac{x-1}{x}}\ \textgreater \ 0} \right. \\\\ \left \{ {{x\in (-\infty ,2)\cup (3,+\infty )} \atop {x\in (-\infty ,0)\cup (1,+\infty )}} \right. \; \; \to \; \; x\in (-\infty ,0)\cup (1,2)\cup (3,+\infty )$