Решебник по алгебре 9 класс Макарычев Задание 373

 

\[\boxed{\text{373\ (373).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 2 \cdot \left( x^{2} + \frac{1}{x^{2}} \right) - \left( x + \frac{1}{x} \right) = 2\]

\[Пусть\ \ t = x + \frac{1}{x};\ \ \]

\[t^{2} = \left( x + \frac{1}{x} \right)^{2} = x^{2} + 2 + \frac{1}{x^{2}};\ \]

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

\[2 \cdot \left( t^{2} - 2 \right) - t = 2\]

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

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

\[D = 1 + 4 \cdot 2 \cdot 6 = 49\]

\[t_{1,2} = \frac{1 \pm 7}{4} = 2;\ - \frac{3}{2}\text{.\ }\]

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

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

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

\[x = 1.\]

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

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

\[D = 9 - 4 \cdot 2 \cdot 2 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:x = 1.\]

\[\textbf{б)}\ \ 9x^{2} - 18x + \frac{9}{x^{2}} - \frac{18}{x} = 22\]

\[9 \cdot \left( x^{2} + \frac{1}{x^{2}} \right) - 18 \cdot \left( x + \frac{1}{x} \right) = 22.\]

\[Пусть\ \ \ t = x + \frac{1}{x};\ \ \]

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

\[9 \cdot \left( t^{2} - 2 \right) - 18t = 22\]

\[9t^{2} - 18 - 18t - 22 = 0\]

\[9t^{2} - 18t - 40 = 0\]

\[D_{1} = 81 + 9 \cdot 40 = 441\]

\[t_{1,2} = \frac{9 \pm 21}{9} = - \frac{4}{3};\frac{10}{3}\text{.\ }\]

\[1)\ x + \frac{1}{x} = \frac{10}{3}\]

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

\[D = 25 - 9 = 16\]

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

\[2)\ x + \frac{1}{x} = - \frac{4}{3}\ \]

\[3x^{2} - 4x + 3 = 0\]

\[D = 4 - 3 \cdot 3 < 0 \Longrightarrow корней\ нет.\]

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

\[\boxed{\text{373.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

Пояснение.

Решение.

\[(x - 3)^{2} + (y - 8)^{2} = r^{2}.\]

\[\textbf{а)}\ точка\ касания\ (3;0):\ \ \ \]

\[(3 - 3)^{2} + (0 - 8)^{2} = 64;\]

\[(x - 3)^{2} + (y - 8)^{2} = 64.\]

\[\textbf{б)}\ точка\ касания\ (0;8):\ \ \]

\[(0 - 3)^{2} + (8 - 8)^{2} = 9;\]

\[(x - 3)^{2} + (y - 8)^{2} = 9.\]