Question

Решите систему неравенств:

Answer 1

$\displaystyle\bf\\\left \{ {{ \frac{2x - 1}{x} < 0} \atop { \frac{3x + 5}{x - 2} \leqslant 0 }} \right. \\ \\ 1) \: \frac{2x - 1}{x} < 0 \\ x(2x - 1) < 0 \\ x(x - 0.5) < 0 \\ + + + (0) - - - (0.5) + + + \\ 0 < x < 0.5 \\ \\ 2) \: \frac{3x + 5}{x - 2} \leqslant 0 \\ \frac{x + 1 \frac{2}{5} }{x - 2} \leqslant 0 \\ \displaystyle\bf\\\left \{ {{(x + 1 \frac{2}{3})(x - 2) \leqslant 0 } \atop {x\neq2 }} \right. \\ \\ + + + + [ - 1 \frac{2}{3} ] - - - - (2) + + + + \\ - 1 \frac{2}{3} \leqslant x < 2 \\ \displaystyle\bf\\3) \: \left \{ {{0 < x < 0.5} \atop { - 1 \frac{2}{3} \leqslant x < 2}} \right. \\ \\ x \: \epsilon \: (0; \: 0.5)$