Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1606

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\[1)\ \left\{ \begin{matrix} x - 3y = - 5\ \ \ \ \ \\ \frac{x}{3y} - \frac{2y}{x} = - \frac{23}{6} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = 3y - 5\ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{x}{3y} - \frac{2y}{x} + \frac{23}{6} = 0 \\ \end{matrix} \right.\ \]

\[225y^{2} - 525y + 150 = 0\ \ \ \ \ \ |\ :75\]

\[3y^{2} - 7y + 2 = 0\]

\[D = 49 - 24 = 25\]

\[y_{1} = \frac{7 - 5}{2 \bullet 3} = \frac{2}{6} = \frac{1}{3};\]

\[y_{2} = \frac{7 + 5}{2 \bullet 3} = \frac{12}{6} = 2;\]

\[x_{1} = 3 \bullet \frac{1}{3} - 5 = 1 - 5 = - 4;\]

\[x_{2} = 3 \bullet 2 - 5 = 6 - 5 = 1.\]

\[Ответ:\ \ \left( - 4;\ \frac{1}{3} \right);\ \ (1;\ 2).\]

\[2)\ \left\{ \begin{matrix} \frac{x + y}{x - y} + \frac{x - y}{x + y} = \frac{10}{3} \\ x^{2} + y^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \frac{(x + y)^{2} + (x - y)^{2}}{(x - y)(x + y)} = \frac{10}{3} \\ x^{2} = 5 - y^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{x^{2} + 2xy + y^{2} + x^{2} - 2xy + y^{2}}{x^{2} - y^{2}} = \frac{10}{3} \\ x^{2} = 5 - y^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \frac{2\left( x^{2} + y^{2} \right)}{x^{2} - y^{2}} = \frac{10}{3} \\ x^{2} = 5 - y^{2}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\frac{2\left( 5 - y^{2} + y^{2} \right)}{5 - y^{2} - y^{2}} = \frac{10}{3}\]

\[\frac{10}{5 - 2y^{2}} = \frac{10}{3}\]

\[5 - 2y^{2} = 3\]

\[2y^{2} = 2\]

\[y^{2} = 1\]

\[y = \pm 1;\]

\[x = \pm \sqrt{5 - ( \pm 1)^{2}} = \pm \sqrt{5 - 1} =\]

\[= \pm \sqrt{4} = \pm 2.\]

\[Ответ:\ \ ( - 2;\ - 1);\ \ ( - 2;\ 1);\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }(2;\ - 1);\ \ (2;\ 1).\]