Решебник по алгебре 9 класс Макарычев Задание 279

 

\[\boxed{\text{279}\text{\ (279)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

\[= 625 - 576 = 49\]

\[t_{1} = \frac{25 + 7}{2} = 16;\ \ \]

\[t_{2} = \frac{25 - 7}{2} = 9;\]

\[\left\{ \begin{matrix} x^{2} = 16 \\ x^{2} = 9\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 4 \\ x = \pm 3 \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm 3;\ \ x = \pm 4.\]

\[\textbf{б)}\ y^{4} + 14y^{2} + 48 = 0\]

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

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

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

\[t_{1} = - 7 + 1 = - 6;\ \]

\[t_{2} = - 7 - 1 = - 8.\]

\[Ответ:исходное\ уравнение\ \]

\[не\ имеет\ корней.\]

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

\[Пусть\ x^{2} = t;\ \ \ t \geq 0:\]

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

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

\[t - 2 = 0\]

\[t = 2.\]

\[x^{2} = 2\]

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

\[Ответ:x = \pm \sqrt{2}.\]

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

\[Пусть\ t^{2} = a;\ \ a \geq 0:\]

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

\[D_{1} = 1 + 3 = 4\]

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

\[t^{2} = 3\]

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

\[Ответ:t = \pm \sqrt{3}.\]

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

\[Пусть\ x^{2} = t;\ \ t \geq 0:\]

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

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

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

\[\ t_{2} = \frac{9 + 7}{4} = 4;\]

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

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

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

\[Пусть\ y^{2} = t;\ \ t \geq 0:\]

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

\[D = 25 - 4 \cdot 5 \cdot 2 =\]

\[= - 15 < 0 \Longrightarrow корней\ нет.\ \]

\[Ответ:нет\ корней.\]

\[\boxed{\text{279.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[Пусть\ \ x - ширина\ \]

\[прямоугольника,\ \]

\[(x + 5) - его\ длина.\]

\[Площадь\ S = x(x + 5);\ \ \]

\[S > 36 - по\ условию.\]

\[Составим\ неравенство:\]

\[x(x + 5) > 36\]

\[x^{2} + 5x - 36 > 0\]

\[x_{1} + x_{2} = - 5;\ \ \ x_{1} \cdot x_{2} = - 36\]

\[x_{1} = 4;\ \ x_{2} = - 9;\]

\[(x + 9)(x - 4) > 0\]

\[x \in ( - \infty;9) \cup (4;\ + \infty).\]

\[Но\ так\ как\ x - сторона\ \]

\[прямоугольника,\ то\ x > 0 \Longrightarrow\]

\[\Longrightarrow x \in (4; + \infty).\]

\[Ответ:больше\ 4\ см.\]