Question

Алгебра 9-й клас. Завдання на фото:

Answer 1

ответы смотри на фото

Answer 2

$\displaystyle\bf\\1)\\\\\frac{4x}{3} +\frac{x}{2} < 11\\\\\\\frac{4x}{3} \cdot 6+\frac{x}{2} \cdot 6 < 11\cdot 6\\\\\\4x\cdot 2+x\cdot 3 < 66\\\\8x+3x < 66\\\\11x < 66\\\\x < 6\\\\\boxed{x\in(-\infty \ ; \ 6)}\\\\2)\\\\\frac{2x}{3} -\frac{3x}{4} \geq \frac{1}{6} \\\\\\\frac{2x}{3}\cdot 12 -\frac{3x}{4} \cdot 12\geq \frac{1}{6} \cdot 12\\\\\\2x\cdot 4-3x\cdot 3\geq 2\\\\8x-9x\geq 2\\\\-x\geq 2\\\\x\leq -2\\\\\boxed{x\in\Big(-\infty \ ; \ -2\Big]}$

$\displaystyle\bf\\3)\\\\\frac{5x}{7} -x > -4\\\\\\\frac{5x}{7} \cdot 7-x\cdot 7 > -4\cdot 7\\\\\\5x-7x > -28\\\\-2x > -28\\\\x < 14\\\\\boxed{x\in(-\infty \ ; \ 14)}\\\\4)\\\\\frac{x}{8} -\frac{1}{4}\leq x\\\\\\\frac{x}{8} \cdot 8-\frac{1}{4}\cdot 8\leq x\cdot 8\\\\\\x-2\leq 8x\\\\x-8x\leq 2\\\\-7x\leq 2\\\\x\geq -\frac{2}{7} \\\\\boxed{x\in\Big[-\frac{2}{7} \ ; \ +\infty\Big)}$