Решебник по алгебре 9 класс Макарычев Задание 370

 

\[\boxed{\text{370\ (370).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{x^{2} + 1}{x} + \frac{x}{x^{2} + 1} = 2\frac{1}{2}\]

\[Пусть\ \ \ \ \ t = \frac{x^{2} + 1}{x}:\text{\ \ }\]

\[t^{\backslash t} + \frac{1}{t} = 2\frac{1}{2}\ \]

\[\frac{t^{2} + 1}{t} = \frac{5}{2}\text{\ \ }\]

\[2t^{2} + 2 = 5t,\ \]

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

\[D = 25 - 4 \cdot 2 \cdot 2 = 9\]

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

\[1)\ \frac{x^{2} + 1}{x} = 2\ \ \]

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

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

\[x - 1 = 0\]

\[x_{1} = 1.\]

\[2)\ \frac{x^{2} + 1}{x} = \frac{1}{2}\text{\ \ }\]

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

\[D = 1 - 4 \cdot 2 \cdot 2 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:x = 1.\]

\[\textbf{б)}\ \frac{x^{2} + 2}{3x - 2} + \frac{3x - 2}{x^{2} + 2} = 2\frac{1}{6}\]

\[Пусть\ \ t = \frac{x^{2} + 2}{3x - 2}:\]

\[t + \frac{1}{t} = 2\frac{1}{6}\]

\[\frac{t^{2} + 1}{t} = \frac{13}{6}\text{\ \ }\]

\[6t^{2} - 13t + 6 = 0,\]

\[D = 13^{2} - 4 \cdot 6 \cdot 6 = 25\]

\[t_{1,2} = \frac{13 \pm 5}{12} = \frac{3}{2};\ \ \ \frac{2}{3}.\]

\[1)\ \frac{x^{2} + 2}{3x - 2} = \frac{3}{2}\text{\ \ }\]

\[2x^{2} + 4 = 9x - 6\ \ \]

\[2x^{2} - 9x + 10 = 0\]

\[D = 81 - 4 \cdot 2 \cdot 10 = 1\]

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

\[x_{2} = \frac{9 + 1}{4} = 2,5;\]

\[2)\ \frac{x^{2} + 2}{3x - 2} = \frac{2}{3}\]

\[3x^{2} + 6 = 6x - 4\]

\[3x^{2} - 6x + 10 = 0\]

\[D = 9 - 3 \cdot 10 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:\ \ x = \ 2;\ \ x = 2,5.\]

\[\boxed{\text{370.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\frac{(2x + y)^{2}}{4} - (x - 0,5y)^{2} = 24\]

\[x^{2} + xy + \frac{y^{2}}{4} - x^{2} - \frac{y^{2}}{4} + xy =\]

\[= 24\]

\[xy = 12\ \ \]

\[y = \frac{12}{x} \Longrightarrow гипербола.\]

\[Ответ:3.\]