Решебник по алгебре 9 класс Мерзляк Задание 690

 

\[\boxed{\mathbf{690\ (690).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ y = \sqrt{3x - 2x^{2}}\]

\[3x - 2x^{2} \geq 0\]

\[x(3 - 2x) \geq 0\]

\[Ответ:x \in \lbrack 0;1,5\rbrack.\]

\[2)\ y = \sqrt{\frac{x - 5}{x + 7}}\]

\[\left\{ \begin{matrix} \frac{x - 5}{x + 7} \geq 0 \\ x \neq - 7\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x \in ( - \infty;\ - 7) \cup \lbrack 5;\ + \infty).\]

\[\boxed{\mathbf{690.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[2x + 1,\ x + 5,\ x + 11\]

\[(x + 5)^{2} = (2x + 1)(x + 11)\ \]

\[x^{2} + 10x + 25 = 2x^{2} + 22x +\]

\[+ x + 11\]

\[x^{2} + 13x - 14 = 0\]

\[x_{1} + x_{2} = - 13;\ \ x_{1} = - 14\]

\[x_{1}x_{2} = - 14;\ \ x_{2} = 1\]

\[при\ x = 1:\ \ \]

\[2 \cdot 1 + 1 = 3,\ \ 1 + 5 = 6,\ \]

\[\ 1 + 11 = 12;\]

\[при\ x = - 14:\ \]

\[\ 2 \cdot ( - 14) + 1 = - 27,\]

\[\ \ - 14 + 5 = - 9,\ \ \]

\[- 14 + 11 = - 3.\]

\[Ответ:x = 1 \Longrightarrow b = 3;6;12\ \ или\ \]

\[\text{\ \ }x = - 14 \Longrightarrow b = - 27;\ - 9;\ - 3.\]