Question

Срочно! Дам 40 баллов!
номер 977 решите системы неравенств
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Answer 1

1.

$\displaystyle\bf\\\left \{ {{ \frac{x}{6} + \frac{x}{3} < 2 \: \: | \times 6 } \atop {2 - \frac{1}{3} x > 0}} \right. \\ \displaystyle\bf\\\left \{ {{x + 2x < 12} \atop { - \frac{1}{3} x > - 2 \: \: | \times ( - 3) }} \right. \\ \displaystyle\bf\\\left \{ {{3x <1 2 \: \: | \div 3} \atop {x < 6 }} \right. \\\displaystyle\bf\\\left \{ {{x < 4} \atop {x < 6 }} \right. \\ \\ x \: \in \: ( - \infty \: ; \: 4 )$

2.

$\displaystyle\bf\\\left \{ {{x - \frac{x + 3}{2} \geqslant 1 \: \: | \times 2 } \atop { - \frac{x}{2} \leqslant 2 - \frac{x}{3} \: \: | \times 6 }} \right. \\ \displaystyle\bf\\\left \{ {{2x - x - 3 \geqslant 2} \atop { - 3x \leqslant 2 - 2x }} \right. \\ \displaystyle\bf\\\left \{ {{x \geqslant 2 + 3} \atop { - 3x + 2x \leqslant 2 }} \right. \\ \displaystyle\bf\\\left \{ {{x \geqslant 5} \atop { - x \leqslant 2 \: \: | \times ( - 1)}} \right. \\ \displaystyle\bf\\\left \{ {{x \geqslant 5} \atop {x \geqslant - 2 }} \right. \\ \\ x \: \in \: [5 \: ; \: + \infty )$