Решебник по алгебре 9 класс Мерзляк Задание 132

 

\[\boxed{\text{132\ (132).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ \frac{4x}{3} + \frac{x}{2} < 11\ \ \ \ | \cdot 6\]

\[8x + 3x < 66\]

\[11x < 66\]

\[x < 6\]

\[Ответ:x \in ( - \infty;6).\]

\[2)\frac{2x}{3} - \frac{3x}{4} \geq \frac{1}{6}\ \ \ \ \ \ \ | \cdot 12\]

\[8x - 9x \geq 2\]

\[- x \geq 2\]

\[x \leq - 2\]

\[Ответ:x \in ( - \infty;2\rbrack.\]

\[3)\frac{5x}{7} - x > - 4\ \ \ \ \ \ \ \ \ | \cdot 7\]

\[- 2x > - 28\]

\[x < 14\]

\[Ответ:x \in ( - \infty;14).\]

\[4)\ \frac{x}{8} - \frac{1}{4} \leq x\ \ \ \ \ \ \ \ | \cdot 8\]

\[x - 2 \leq 8x\]

\[- 7x \leq 2\]

\[x \geq - \frac{2}{7}\]

\[Ответ:x \in \left\lbrack - \frac{2}{7};\ + \infty \right).\]

\[\boxed{\mathbf{132.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 3x^{2} + 4x - 20 = 0\]

\[D_{1} = 4 + 60 = 64\]

\[x_{1} = \frac{- 2 + 8}{3} = \frac{6}{3} = 2;\]

\[x_{2} = \frac{- 2 - 8}{3} = - \frac{10}{3} = - 3\frac{1}{3}.\]

\[2)\ x^{2} - 6x + 12 = 5x - x^{2}\]

\[2x^{2} - 11x + 12 = 0\]

\[D = 121 - 96 = 25\]

\[x_{1} = \frac{11 + 5}{4} = 4;\]

\[x_{2} = \frac{11 - 5}{4} = \frac{6}{4} = 1,5.\]

\[3)\ (x - 3)^{2} = 8 - x - 3x^{2}\]

\[x^{2} - 6x + 9 = 8 - x - 3x^{2}\]

\[4x^{2} - 5x + 1 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 + 3}{8} = 1;\]

\[x_{2} = \frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{4} = 0,25.\]

\[4)\ (x + 1)(x - 5) = 2\]

\[x^{2} + x - 5x - 5 - 2 = 0\]

\[x^{2} - 4x - 7 = 0\]

\[D_{1} = 4 + 7 = 11\]

\[x = 2 \pm \sqrt{11}.\]