Решебник по алгебре 9 класс Мерзляк Задание 195

 

\[\boxed{\text{195\ (195).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\left\{ \begin{matrix} 4x + 3 \geq 6x - 7\ \ \ \ \ \ \ \ \\ 3(x + 8) \geq 4(8 - x) \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 2x \geq - 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x + 24 \geq 32 - 4x \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x \leq 5\ \ \\ 7x \geq 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x \leq 5\ \ \ \\ x \geq 1\frac{1}{7} \\ \end{matrix} \right.\ \]

\[x \in \left\lbrack 1\frac{1}{7};5 \right\rbrack\]

\[Целые\ решения:2;3;4;5.\]

\[Ответ:4.\]

\[2)\ \left\{ \begin{matrix} x - \frac{x + 1}{3} - \frac{x - 2}{6} < 2\ \ | \cdot 6 \\ \frac{2x - 5}{3} \geq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 6x - 2x - 2 - x + 2 < 12 \\ 2x - 5 \geq - 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3x < 12 \\ 2x \geq - 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x < 4\ \ \ \\ x \geq - 2 \\ \end{matrix} \right.\ \]

\[x \in \lbrack - 2;4)\]

\[Целые\ решения:\ - 2;\ - 1;0;1;\]

\[2;3.\]

\[Ответ:6.\]

\[\boxed{\text{195.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{2} + 6x - 2 = 0\]

\[D = 36 + 8 = 44\]

\[x_{1,2} = \frac{- 6 \pm \sqrt{44}}{2}\]

\[x_{1} = \frac{- 6 + 2\sqrt{11}}{2} = - 3 + \sqrt{11}\]

\[x_{2} = \frac{- 6 - 2\sqrt{11}\ }{2} = - 3 - \sqrt{11}\]

\[x_{1}^{2} + x_{2}^{2} = \left( - 3 + \sqrt{11} \right)^{2} +\]

\[+ \left( - 3 - \sqrt{11} \right)^{2} =\]

\[= 9 - 6\sqrt{11} + 11 + 9 +\]

\[+ 6\sqrt{11} + 11 = 40\]

\[Ответ:40.\]