Решебник по алгебре 9 класс Макарычев Задание 1016

 

\[\boxed{\text{1016\ (1016).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \sqrt{12x - 4}\]

\[12x - 4 \geq 0\]

\[12x \geq 4\]

\[x \geq \frac{1}{3}.\]

\[Ответ:при\ x \geq \frac{1}{3}.\]

\[\textbf{б)}\ \sqrt{3 - 0,6x}\]

\[3 - 0,6x \geq 0\]

\[- 0,6x \geq - 3\]

\[x \leq 5.\]

\[Ответ:при\ x \leq 5.\]

\[\textbf{в)}\ \sqrt{15 + 2x - x²}\]

\[15 + 2x - x^{2} \geq 0\]

\[x^{2} - 2x - 15 \leq 0\]

\[D = 4 + 60 = 64,\]

\[x_{1} = \frac{2 + 8}{2} = 5,\ \ \]

\[x_{2} = \frac{2 - 8}{2} = - 3,\]

\[- 3 \leq x \leq 5.\]

\[Ответ:при\ \ \ - 3 \leq x \leq 5.\]

\[\textbf{г)}\ \sqrt{2x² + x - 6}\]

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

\[D = 1 + 48 = 49,\]

\[x_{1} = \frac{- 1 + 7}{4} = 1,5,\ \ \]

\[x_{2} = \frac{- 1 - 7}{4} = - 2,\]

\[x \in ( - \infty;\ - 2\rbrack \cup \lbrack 1,5;\ + \infty).\]

\[Ответ:\]

\[при\ \ x \in ( - \infty;\ - 2\rbrack \cup \lbrack 1,5;\ + \infty).\]

\[\textbf{д)}\ \sqrt{12 - 5x} + \sqrt{2x - 1}\]

\[0,5 \leq x \leq 2,4.\]

\[Ответ:при\ \ \ 0,5 \leq x \leq 2,4.\]

\[\textbf{е)}\ \sqrt{x² + 4} + \sqrt{3x - 17}\]

\[x \geq 5\frac{2}{3}.\]

\[Ответ:при\ x \geq 5\frac{2}{3}.\]