ГДЗ по алгебре 9 класс Макарычев Задание 282

 

\[\boxed{\text{282\ (282).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( x^{2} - 1 \right)\left( x^{2} + 1 \right) -\]

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

\[x^{4} - 1 - 4x^{2} + 44 = 0\]

\[x^{4} - 4x^{2} + 43 = 0\]

\[Пусть\ x^{2} = t;\ \ x^{4} = t^{2};\ \ t \geq 0:\]

\[t^{2} - 4t + 43 = 0\]

\[D_{1} = 4 - 43 = - 39 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:нет\ корней.\]

\[\textbf{б)}\ 3x^{2}(x - 1)(x + 1) - 10x^{2} +\]

\[+ 4 = 0\]

\[3x^{2}\left( x^{2} - 1 \right) - 10x^{2} + 4 = 0\]

\[3x^{4} - 3x^{2} - 10x^{2} + 4 = 0\]

\[3x^{4} - 13x^{2} + 4 = 0\]

\[Пусть\ x^{2} = t;\ \ x^{4} = t^{2};\ \ t \geq 0:\]

\[3t^{2} - 13t + 4 = 0\]

\[D = 13^{2} - 4 \cdot 3 \cdot 4 =\]

\[= 169 - 48 = 121\]

\[t_{1} = \frac{13 + 11}{6} = 4;\ \]

\[\ t_{2} = \frac{13 - 11}{6} = \frac{1}{3}.\]

\[\left\{ \begin{matrix} x^{2} = 4 \\ x^{2} = \frac{1}{3}\ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 2\ \ \ \\ x = \pm \sqrt{\frac{1}{3}} \\ \end{matrix} \right.\ .\]

\[Ответ:x = \pm 2;\ \ x = \pm \sqrt{\frac{1}{3}}.\]

\[\boxed{\text{282.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[y = \frac{0,5x - 2}{3};\ \ y = \frac{1}{6}x - \frac{2}{3};\]

\[Функция\ возрастает,\ так\ как\ \]

\[k = \frac{1}{6} > 0.\]

\[Точка\ пересечения\ с\ Ox:\]

\[y = 0 \Longrightarrow \frac{0,5x - 2}{3} = 0 \Longrightarrow\]

\[\Longrightarrow 0,5x = 2 \Longrightarrow x = 4 \Longrightarrow\]

\[\Longrightarrow точка\ (4;0).\]

\[Точка\ пересечения\ с\ \text{Oy}:\]

\[x = 0 \Longrightarrow y = - \frac{2}{3} \Longrightarrow\]

\[\Longrightarrow точка\ \left( 0;\ - \frac{2}{3} \right).\]