Решебник по алгебре 9 класс Макарычев Задание 332

 

\[\boxed{\text{332\ (332).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\mathbf{Выражение\ под\ знаком\ корня\ }\]

\[\mathbf{должно\ быть\ }\]

\[\mathbf{неотрицательным\ }\left( \mathbf{\geq 0} \right)\mathbf{.}\]

\[\textbf{а)}\ y = \sqrt{(5 - x)(x + 8)}\]

\[(5 - x)(x + 8) \geq 0\]

\[(x + 8)(x - 5) \leq 0\]

\[x \in \lbrack - 8;5\rbrack.\]

\[\textbf{б)}\ y = \sqrt{(x + 12)(x - 1)(x - 9)}\]

\[(x + 12)(x - 1)(x - 9) \geq 0\]

\[x \in \lbrack - 12;1\rbrack \cup \lbrack 9;\ + \infty).\ \]

\[\boxed{\text{332.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x² = \frac{7x - 4}{4x - 7}\]

\[4x^{3} - 7x^{2} = 7x - 4\]

\[4x^{3} - 7x^{2} - 7x + 4 = 0\]

\[(x + 1)\left( 4x^{2} - 11x + 4 \right) = 0\]

\[1)\ x + 1 = 0,\ \ x_{1} = - 1;\]

\[2)\ 4x² - 11x + 4 = 0\]

\[D = 121 - 4 \cdot 4 \cdot 4 = 57\]

\[x_{1,2} = \frac{11 \pm \sqrt{57}}{8};\]

\[\textbf{б)}\ x² = \frac{5x - 3}{3x - 5}\]

\[3x^{3} - 5x^{2} = 5x - 3\]

\[3x^{3} - 5x^{2} - 5x + 3 = 0\]

\[(x + 1)\left( 3x^{2} - 8x + 3 \right) = 0\]

\[1)\ x + 1 = 0,\ \ x_{1} = - 1;\]

\[2)\ 3x² - 8x + 3 = 0\]

\[D = 16 - 3 \cdot 3 = 7\]

\[x_{1,2} = \frac{4 \pm \sqrt{7}}{3}.\]

\[Ответ:а) - 1;\ \frac{11 \pm \sqrt{57}}{8};\]

\[\textbf{б)} - 1;\frac{4 \pm \sqrt{7}}{3}.\]