Решебник по алгебре 9 класс Мерзляк Задание 152

 

\[\boxed{\text{152\ (152).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ \sqrt{9 - x} + \frac{10}{x + 3}\]

\[9 - x \geq 0\ \ \ \ \ \ и\ \ \ \ \ \ \ \ x + 3 \neq 0\]

\[x \leq 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq - 3\ \]

\[Ответ:( - \infty;9\rbrack;\ кроме - 3.\]

\[\boxed{\mathbf{152.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

\[x^{3} - 1 - 2x(x - 1) = 0\]

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

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

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

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

\[D = 1 - 4 < 0\]

\[нет\ корней.\]

\[x - 1 = 0\]

\[x = 1.\]

\[Ответ:x = 1.\]

\[2)\ x^{4} - x^{2} - 8x - 16 = 0\]

\[x^{4} - \left( x^{2} + 8x + 16 \right) = 0\]

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

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

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

\[D = 1 + 16 = 17\]

\[x = 1 \pm \sqrt{17}.\]

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

\[D = 1 - 16 = - 15 < 0\]

\[нет\ корней.\]

\[Ответ:x = 1 \pm \sqrt{17}.\]

\[3)\ x^{3} - 5x^{2} + 4 = 0\]

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

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

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

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

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

\[D_{1} = 4 + 4 = 8\]

\[x = 2 \pm \sqrt{8} = 2 \pm 2\sqrt{2}.\]

\[x - 1 = 0\]

\[x = 1.\]

\[Ответ:x = 2 \pm 2\sqrt{2};\ \ x = 1.\]

\[4)\ x^{3} - 6x + 5 = 0\]

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

\[x\left( x^{2} - 1 \right) - 5(x - 1) = 0\]

\[x(x - 1)(x + 1) - 5(x - 1) = 0\]

\[(x - 1)\left( x^{2} + x - 5 \right) = 0\]

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

\[D = 1 + 20 = 21\]

\[x = \frac{- 1 \pm \sqrt{21}}{2}.\]

\[x - 1 = 0\]

\[x = 1.\]

\[Ответ:\ x = 1;\ \ x = \frac{- 1 \pm \sqrt{21}}{2}\text{.\ }\]