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МатематикаРешебник по алгебре 9 класс Макарычев Задание 278
Задание 278
\[\boxed{\text{278\ (278).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x^{4} - 5x^{2} - 36 = 0\]
\[Пусть\ \ x^{2} = t;\ \ \ x^{4} = t^{2};\ \ t \geq 0:\]
\[t^{2} - 5t - 36 = 0\]
\[t_{1} + t_{2} = 5;\ \ \ t_{1} \cdot t_{2} = - 36\]
\[t_{1} = 9;\ \ \ \ \ t_{2} = - 4 \Longrightarrow не\]
\[\ подходит.\]
\[x^{2} = 9\ \]
\[x = \pm 3.\]
\[Ответ:x = \pm 3.\]
\[\textbf{б)}\ y^{4} - 6y^{2} + 8 = 0\]
\[Пусть\ \ y^{2} = t;\ \ y^{4} = t^{2};\ \]
\[\ t \geq 0.\]
\[t^{2} - 6t + 8 = 0\]
\[D_{1} = 9 - 8 = 1\]
\[t_{1} = 3 + 1 = 4;\ \ \ \ \ \]
\[t_{2} = 3 - 1 = 2;\]
\[\left\{ \begin{matrix} y^{2} = 4 \\ y^{2} = 2 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{1,2} = \pm 2 \\ y_{3,4} = \pm \sqrt{2} \\ \end{matrix} \right.\ .\]
\[Ответ:y = \pm 2;\ \ y = \pm \sqrt{2}.\]
\[\textbf{в)}\ t^{4} + 10t^{2} + 25 = 0\]
\[Пусть\ t^{2} = a;\ \ t^{4} = a^{2};\ \ a \geq 0.\]
\[a^{2} + 10a + 25 = 0\]
\[(a + 5)^{2} = 0\]
\[a + 5 = 0\]
\[a = - 5 \Longrightarrow не\ подходит.\]
\[Ответ:корней\ исходное\ \]
\[уравнение\ не\ имеет.\]
\[\textbf{г)}\ 4x^{4} - 5x^{2} + 1 = 0\]
\[Пусть\ x^{2} = a;\ \ x^{4} = a^{2};\ \ a \geq 0.\]
\[4a^{2} - 5a + 1 = 0\]
\[D = 25 - 16 = 9\]
\[a_{1} = \frac{5 + 3}{8} = 1;\ \ \ a_{2} = \frac{5 - 3}{8} = \frac{1}{4}.\]
\[\left\{ \begin{matrix} x^{2} = 1 \\ x^{2} = \frac{1}{4} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{1,2} = \pm 1 \\ x_{3,4} = \pm \frac{1}{2} \\ \end{matrix} \right.\ \]
\[Ответ:x = \pm 1;\ \ x = \pm \frac{1}{2}.\]
\[\textbf{д)}\ 9x^{4} - 9x^{2} + 2 = 0\]
\[Пусть\ \ x^{2} = t;\ \ t \geq 0.\]
\[9t^{2} - 9t + 2 = 0\]
\[D = 81 - 72 = 9\]
\[t_{1} = \frac{9 - 3}{18} = \frac{1}{3};\ \ t_{2} = \frac{9 + 3}{18} = \frac{2}{3}.\]
\[x^{2} = \frac{1}{3} \Longrightarrow x = \pm \sqrt{\frac{1}{3}}.\]
\[x^{2} = \frac{2}{3} \Longrightarrow x = \pm \sqrt{\frac{2}{3}}.\]
\[Ответ:\ x = \pm \sqrt{\frac{2}{3}};\ \ x = \pm \sqrt{\frac{1}{3}}.\]
\[\textbf{е)}\ 16y^{4} - 8y^{2} + 1 = 0\]
\[Пусть\ \ y^{2} = t;\ \ \ t \geq 0.\]
\[16t^{2} - 8t + 1 = 0\]
\[(4t - 1)^{2} = 0\]
\[4t - 1 = 0\]
\[4t = 1\]
\[t = \frac{1}{4}.\]
\[y^{2} = \frac{1}{4}\]
\[y = \pm \frac{1}{2}.\]
\[Ответ:y = \pm \frac{1}{2}.\]
\[\boxed{\text{278.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[Пусть\ x - меньшая\ \ сторона\ \]
\[прямоугольника,\ \]
\[тогда\ (x + 7) - его\ большая\ \]
\[сторона.\]
\[Площадь\ S = x(x + 7);\ S \leq 60\ \]
\[по\ условию.\]
\[Составим\ неравенство:\]
\[x(x + 7) \leq 60\]
\[x^{2} + 7x - 60 \leq 0\]
\[D = 49 + 4 \cdot 60 = 289\]
\[x_{1} = \frac{- 7 + 17}{2} = 5;\ \ \ \]
\[x_{2} = \frac{- 7 - 17}{2} = - 12;\]
\[(x + 12)(x - 5) \leq 0\]
\[x \in \lbrack - 12;5\rbrack.\]
\[Так\ как\ x - сторона\ \]
\[прямоугольника,\ то\ x > 0.\]
\[Следовательно,\ x \in (0;5\rbrack\text{.\ }\]
\[Ответ:меньшая\ сторона\ \]
\[больше\ 0,\ но\ меньше\ 5.\]
