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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 134
Задание 134
\[\boxed{\text{134\ (134).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ 3 - 5(2x + 4) \geq 7 - 2x\]
\[3 - 10x - 20 \geq 7 - 2x\]
\[- 10x + 2x \geq 7 - 3 + 20\]
\[- 8x \geq 24\]
\[x \leq - 3\]
\[Ответ:x \in ( - \infty;\ - 3\rbrack.\]
\[2)\ 6x - 3(x - 1) \leq 2 + 5x\]
\[6x - 3x + 3 - 2 - 5x \leq 0\]
\[- 2x \leq - 1\]
\[x \geq 0,5\]
\[Ответ:x \in \lbrack 0,5;\ + \infty).\]
\[3)\ x - 2(x - 1) \geq 10 + 3(x + 4)\]
\[x - 2x + 2 \geq 10 + 3x + 12\]
\[x - 2x - 3x \geq 22 - 2\]
\[- 4x \geq 20\]
\[x \leq - 5\]
\[Ответ:x \in ( - \infty;\ - 5\rbrack.\]
\[4)\ 2(2x - 3,5) -\]
\[- 3(2 - 3x) < 6(1 - x)\]
\[4x - 7 - 6 + 9x < 6 - 6x\]
\[4x + 9x + 6x < 6 + 7 + 6\]
\[19x < 19\]
\[x < 1\]
\[Ответ:x \in ( - \infty;1).\]
\[5)\ (x + 1)(x - 2)\ \leq\]
\[\leq (x - 3)(x + 3)\]
\[x^{2} - 2x + x - 2 \leq x^{2} + 3x -\]
\[- 3x - 9\]
\[x^{2} - 2x + x - x^{2} -\]
\[- 3x + 3x \leq - 9 + 2\]
\[- x \leq - 7\]
\[x \geq 7\]
\[Ответ:x \in \lbrack 7;\ + \infty).\]
\[6)\ (4x - 3)^{2} + (3x + 2)^{2} \geq\]
\[\geq (5x + 1)^{2}\]
\[16x^{2} - 24x + 9 + 9x^{2} + 12x +\]
\[+ 4 \geq 25x^{2} + 10x + 1\ \]
\[16x^{2} + 9x^{2} - 25x^{2} - 24x +\]
\[+ 12x - 10x \geq 1 - 9 - 4\]
\[- 22x \geq - 12\]
\[x \leq \frac{6}{11}\]
\[Ответ:x \in \left( - \infty;\frac{6}{11} \right\rbrack.\]
\[7)\ \frac{2x - 1}{4} \geq \frac{3x - 5}{5}\]
\[\frac{2x - 1}{4} - \frac{3x - 5}{5} \geq 0\]
\[\frac{10x - 5 - 12x + 20}{20} \geq 0\]
\[- 2x + 5 \geq 0\]
\[- 2x \geq - 15\]
\[x \leq 7,5\]
\[Ответ:x \in ( - \infty;7,5\rbrack.\]
\[8)\ \frac{3x + 7}{4} - \frac{5x - 2}{2} < x\]
\[\frac{3x + 7 - 10x + 4}{4} < x\ \ | \cdot 4\]
\[- 7x + 11 < 4x\]
\[- 11x < - 11\]
\[x > 1\]
\[Ответ:x \in (1;\ + \infty).\]
\[9)\ (x - 5)(x + 1) \leq 3 + (x - 2)^{2}\]
\[x^{2} + x - 5x - 5 \leq 3 + x^{2} -\]
\[- 4x + 4\]
\[x^{2} - x^{2} + x - 5x + 4x \leq 3 +\]
\[+ 4 + 5\]
\[0 \leq 12\]
\[Ответ:x \in ( - \infty; + \infty).\]
\[10)\ \frac{x + 1}{2} - \frac{x - 3}{3} > 2 + \frac{x}{6}\]
\[\frac{3x + 3 - 2x + 6}{6} > \frac{12 + x}{6}\ \ \ | \cdot 6\]
\[3x + 3 - 2x + 6 > 12 + x\]
\[x + 9 > 12 + x\]
\[9 > 12 - неверно.\]
\[Ответ:\ \varnothing.\]
\[11)\ (6x - 1)^{2} - 4x(9x - 3) \leq 1\]
\[36x^{2} - 12x + 1 - 36x^{2} +\]
\[+ 12x \leq 1\]
\[1 \leq 1\]
\[Ответ:x \in ( - \infty; + \infty).\]
\[12)\ \frac{x - 3}{9} - \frac{x + 4}{4} > \frac{x - 8}{6}\]
\[\frac{4x - 12 - 9x - 36}{36} > \frac{x - 8}{6}\ | \cdot 36\]
\[- 5x - 48 > 6x - 48\]
\[- 11x > 0\]
\[x < 0\]
\[Ответ:x \in ( - \infty;0).\ \]
\[\boxed{\text{134}\text{\ (}\text{с}\text{)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ \frac{y}{6} - \frac{5y}{4} < 1\ \ \ \ \ \ \ \ \ | \cdot 12\]
\[2y - 15y \leq 12\]
\[- 13y \leq 12\]
\[y \geq - \frac{12}{13}\]
\[Ответ:x \in \left\lbrack - \frac{12}{13};\ + \infty \right).\]
\[2)\ \frac{x}{10} - \frac{x}{5} > - 2\ \ \ \ \ \ \ \ \ \ \ | \cdot 10\]
\[x - 2x > - 20\]
\[- x > - 20\]
\[x < 20\]
\[Ответ:x \in ( - \infty;20)\text{.\ }\]
