Решебник по алгебре 9 класс Макарычев Задание 223

 

\[\boxed{\text{223}\text{\ (223)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[a > 0;\ \ b > 0.\]

\[a + b = 40\]

\[a = 40 - b\]

\[a \cdot b = b(40 - b) = - b^{2} + 40b =\]

\[= - \left( b^{2} - 2 \cdot 20b + 400 - 400 \right) =\]

\[= - \left( (b - 20)^{2} - 400 \right) =\]

\[= 400 - (b - 20)^{2} \leq 400.\]

\[Наибольшее\ значение\ 400\ \]

\[при\ \ (b - 20)^{2} = 0:\]

\[b - 20 = 0\]

\[b = 20.\]

\[\Longrightarrow a = 40 - b = 40 - 20 = 20.\]

\[Ответ:при\ a = 20;b = 20.\]

\[\boxed{\text{223.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x^{4} - 5x^{2} - 36 = 0\]

\[Пусть\ \ x^{2} = t,\ x^{4} = t^{2},\ \ t \geq 0.\]

\[t^{2} - 5t - 36 = 0\]

\[D = 25 + 4 \cdot 36 = 169\]

\[t_{1,2} = \frac{5 \pm 13}{2}\text{\ \ }\]

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

\[\Longrightarrow не\ подходит.\]

\[x^{2} = 9\ \]

\[x = \pm 3.\]

\[Ответ:x = \pm 3.\]

\[\textbf{б)}\ y^{4} - 6y^{2} + 8 = 0\]

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

\[\ \ t \geq 0.\]

\[t^{2} - 6t + 8 = 0\]

\[D = 9 - 8 = 1\]

\[t_{1,2} = 3 \pm 1\]

\[t_{1} = 4;\ \ \ \ \ t_{2} = 2;\]

\[\left\{ \begin{matrix} y² = 4 \\ y² = 2 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{1,2} = \pm 2 \\ y_{3,4} = \pm \sqrt{2} \\ \end{matrix} \right.\ .\]

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

\[\textbf{в)}\ t^{4} + 10t² + 25 = 0\]

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

\[\ \ a \geq 0.\]

\[a^{2} + 10a + 25 = 0\]

\[(a + 5)^{2} = 0\]

\[a = - 5 \Longrightarrow не\ подходит.\]

\[Ответ:корней\ исходное\ \]

\[уравнение\ не\ имеет.\]

\[\textbf{г)}\ 4x^{4} - 5x^{2} + 1 = 0\]

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

\[\ \ a \geq 0.\]

\[4a^{2} - 5a + 1 = 0\]

\[D = 25 - 16 = 9\]

\[a_{1,2} = \frac{5 \pm 3}{8} = 1;\frac{1}{4}.\]

\[\left\{ \begin{matrix} x² = 1 \\ x² = \frac{1}{4} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{1,2} = \pm 1 \\ x_{3,4} = \pm \frac{1}{2} \\ \end{matrix} \right.\ \]

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

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

\[Пусть\ \ x^{2} = t,\ \ t \geq 0.\]

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

\[D = 9\]

\[t_{1,2} = \frac{9 \pm 3}{18} = \frac{1}{3};\ \frac{2}{3}.\]

\[x^{2} = \frac{1}{3} \Longrightarrow x = \pm \sqrt{\frac{1}{3}}.\]

\[x² = \frac{2}{3} \Longrightarrow x = \pm \sqrt{\frac{2}{3}}.\]

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

\[\textbf{е)}\ 16y^{4} - 8y^{2} + 1 = 0\]

\[Пусть\ \ y^{2} = t,\ \ t \geq 0.\]

\[16t^{2} - 8t + 1 = 0\]

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

\[t = \frac{1}{4} \Longrightarrow y^{2} = \frac{1}{4} \Longrightarrow\]

\[y = \pm \frac{1}{2}.\]

\[Ответ:y = \pm \frac{1}{2}.\]