ГДЗ по алгебре 9 класс Мерзляк Задание 135

\[\boxed{\text{135\ (135).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ 3(4x + 9) + 15 > 7(8 - x)\]

\[12x + 27 + 5 > 56 - 7x\]

\[19x > 56 - 32\]

\[19x > 24\]

\[x > \frac{24}{19}\]

\[Ответ:x \in \left( 1\frac{5}{19};\ + \infty \right).\]

\[2)\ (2 - y)(3 + y) \leq\]

\[\leq (4 + y)(6 - y)\ \]

\[6 + 2y - 3y - y^{2} \leq\]

\[\leq 24 - 4y + 6y - y^{2}\]

\[- y - y^{2} - 2y + y^{2} \leq 24 - 6\]

\[- 3y \leq 18\]

\[y \geq - 6\]

\[Ответ:x \in \lbrack - 6;\ + \infty).\]

\[3)\ (y + 3)(y - 5) -\]

\[- (y - 1)^{2} > - 16\]

\[y^{2} - 5y + 3y - 15 -\]

\[- y^{2} + 2y - 1 + 16 > 0\]

\[0 > 0 - неверно.\]

\[Ответ:\ \varnothing.\]

\[4)\ \frac{3x - 7}{5} - 1 \geq \frac{2x - 6}{3}\ \ | \cdot 15\]

\[3 \cdot (3x - 7) - 15 \geq 5 \cdot (2x - 6)\]

\[9x - 21 - 15 - 10x + 30 \geq 0\]

\[- x - 6 \geq 0\]

\[- x \geq 6\]

\[x \leq - 6\]

\[Ответ:x \in ( - \infty; - 6\rbrack.\]

\[5)\ \frac{2x}{3} - \frac{x - 1}{6} - \frac{x + 2}{2} < 0\ | \cdot 6\]

\[4x - x + 1 - 3x - 6 < 0\]

\[0x < 5\]

\[Ответ:x \in ( - \infty;\ + \infty).\]

\[6)\ \frac{y - 1}{2} - \frac{2y + 1}{8} - y < 2\ \ | \cdot 8\]

\[4y - 4 - 2y - 1 - 8y < 16\]

\[- 6y - 5 < 16\]

\[- 6y < 21\]

\[y > - \frac{21}{6}\]

\[y > - 3\frac{3}{6}\]

\[y > - 3,5\]

\[Ответ:y \in ( - 3,5; + \infty).\]