ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 342

\[\boxed{\mathbf{342}\mathbf{.}}\]

\[1)\ \left\{ \begin{matrix} \lg x - \lg y = 2\ \\ x - 10y = 900 \\ \end{matrix} \right.\ \]

\[1)\ \lg x - \lg y = 2\]

\[\lg\frac{x}{y} = \lg 10^{2}\]

\[\frac{x}{y} = 10^{2}\]

\[\frac{x}{y} = 100\]

\[x = 100y.\]

\[2)\ x - 10y = 900\]

\[100y - 10y = 900\]

\[90y = 900\ \]

\[y = 10.\]

\[x = 100 \bullet 10 = 1000.\]

\[Ответ:\ \ (1000;10).\]

\[2)\ \left\{ \begin{matrix} \log_{3}x + \log_{3}y = 2 \\ x^{2}y - 2y + 9 = 0\ \ \\ \end{matrix} \right.\ \]

\[1)\ \log_{3}x + \log_{3}y = 2\]

\[\log_{3}\left( \text{xy} \right) = \log_{3}3^{2}\]

\[xy = 3^{2}\]

\[xy = 9\]

\[y = \frac{9}{x}.\]

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

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

\[9x + 9 - \frac{18}{x} = 0\ \ \ \ \ | \bullet \frac{x}{9}\]

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[x_{1} = \frac{- 1 - 3}{2} = - 2;\text{\ \ }\]

\[x_{2} = \frac{- 1 + 3}{2} = 1.\]

\[Так\ как\ x > 0\ и\ y > 0:\]

\[x = 1\ \ и\ \ y = \frac{9}{1} = 9.\]

\[Ответ:\ \ (1;\ 9).\]