Решебник по алгебре 9 класс Макарычев Задание 224

 

\[\boxed{\text{224\ (224).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 0,8x^{2} - 19,8x - 5\]

\[0,8x^{2} - 19,8x - 5 = 0\ \ \ \ \ \ \ | \cdot 5\]

\[4x^{2} - 99x - 25 = 0\]

\[D = 9801 + 400 = 10\ 201 = 101^{2}\]

\[x_{1} = \frac{99 + 101}{8} = \frac{200}{8} =\]

\[= \frac{100}{4} = 25;\]

\[x_{2} = \frac{99 - 101}{8} = - \frac{2}{8} =\]

\[= - \frac{1}{4} = - 0,25.\]

\[\Longrightarrow 0,8x^{2} - 19,8x - 5 =\]

\[= \frac{4}{5} \cdot (x - 25)\left( x + \frac{1}{4} \right) =\]

\[= (x - 25)(0,8x + 0,2).\]

\[\textbf{б)}\ \ 3,5 - 3\frac{1}{3}x + \frac{2}{3}x^{2}\]

\[\frac{2}{3}x^{2} - \frac{10}{3}x + 3,5 = 0\ \ \ \ \ | \cdot 6\]

\[4x^{2} - 20x + 21 = 0\]

\[D_{1} = 100 - 84 = 16\]

\[x_{1} = \frac{10 + 4}{4} = \frac{14}{4} = \frac{7}{2} = 3,5;\]

\[x_{2} = \frac{10 - 4}{4} = \frac{6}{4} = \frac{3}{2} = 1,5.\]

\[3,5 - 3\frac{1}{3}x + \frac{2}{3}x^{2} =\]

\[= \frac{2}{3} \cdot (x - 3,5)\left( x - \frac{3}{2} \right) =\]

\[= (x - 3,5)\left( \frac{2}{3}x - 1 \right).\]

\[\textbf{в)}\ x^{2} + x\sqrt{2} - 2\]

\[x^{2} + x\sqrt{2} - 2\ \]

\[D = \left( \sqrt{2} \right)^{2} + 4 \cdot 8 = 2 + 8 = 10\]

\[x_{1,2} = \frac{- \sqrt{2} \pm \sqrt{10}}{2};\]

\[x^{2} + x\sqrt{2} - 2 =\]

\[= \left( x + \frac{\sqrt{2} + \sqrt{10}}{2} \right) \cdot\]

\[\cdot \left( x + \frac{\sqrt{2} - \sqrt{10}}{2} \right).\]

\[\textbf{г)}\ x^{2} - x\sqrt{6} + 1\]

\[x^{2} - x\sqrt{6} + 1 = 0\]

\[D = \left( \sqrt{6} \right)^{2} - 4 \cdot 1 = 6 - 4 = 2\]

\[x_{1,2} = \frac{\sqrt{6} \pm \sqrt{2}}{2}\]

\[x^{2} - x\sqrt{6} + 1 =\]

\[= \left( x - \frac{\sqrt{6} + \sqrt{2}}{2} \right) \cdot\]

\[\cdot \left( x - \frac{\sqrt{6} - \sqrt{2}}{2} \right)\text{.\ }\]

\[\boxed{\text{224.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x^{4} - 25x^{2} + 144 = 0\]

\[Пусть\ x^{2} = t,\ \ x^{4} = t^{2},\]

\[\ \ t \geq 0,\ тогда:\]

\[t^{2} - 25t + 144 = 0\]

\[D = 25^{2} - 4 \cdot 144 =\]

\[= 625 - 576 = 49\]

\[t_{1,2} = \frac{25 \pm 7}{2},\ \ t_{1} = 16,\ \ \]

\[t_{2} = 9;\]

\[\left\{ \begin{matrix} x² = 16 \\ x² = 9\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 4 \\ x = \pm 3 \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm 3;\ \ x = \pm 4.\]

\[\textbf{б)}\ y^{4} + 14y² + 48 = 0\]

\[Пусть\ \ y^{2} = t,\ \ y^{4} = t^{2},\ \]

\[\ t \geq 0,\ тогда:\]

\[t^{2} + 14t + 48 = 0\]

\[D = 7^{2} - 48 = 49 - 48 = 1\]

\[t_{1,2} = - 7 \pm 1,\ \ \]

\[t_{1,2} = - 6; - 8 \Longrightarrow корней\ нет.\]

\[Ответ:исходное\ уравнение\ \]

\[не\ имеет\ корней.\]

\[\textbf{в)}\ x^{4} - 4x^{2} + 4 = 0\]

\[Пусть\ x^{2} = t,\ \ \ t \geq 0,\ тогда:\]

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

\[(t - 2)^{2} = 0,\ \ t = 2;\]

\[x^{2} = 2 \Longrightarrow x = \pm \sqrt{2}\]

\[Ответ:x = \pm \sqrt{2}.\]

\[\textbf{г)}\ t^{4} - 2t^{2} - 3 = 0\]

\[Пусть\ t^{2} = a,\ \ a \geq 0,\ тогда:\]

\[a^{2} - 2a - 3 = 0\]

\[D = 1 + 3 = 4\]

\(a_{1,2} = 1 \pm 2 = - 1; - 3;\)

\[t² = 3 \Longrightarrow t = \pm \sqrt{3}.\]

\[Ответ:t = \pm \sqrt{3}.\]

\[\textbf{д)}\ 2x^{4} - 9x^{2} + 4 = 0\]

\[Пусть\ x^{2} = t,\ \ t \geq 0,\ тогда:\]

\[2t^{2} - 9t + 4 = 0\]

\[D = 81 - 2 \cdot 4 \cdot 4 = 49\]

\[t_{1,2} = \frac{9 \pm 7}{4} = \frac{1}{2};4;\]

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

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

\[\textbf{е)}\ 5y^{4} - 5y^{2} + 2 = 0\]

\[Пусть\ y^{2} = t,\ \ t \geq 0,\ тогда:\]

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

\[D = 25 - 4 \cdot 5 \cdot 2 = - 15 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\ \]

\[Ответ:нет\ корней.\]