Решебник по алгебре 8 класс Макарычев ФГОС Задание 1328

\[\boxed{\text{1328.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

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

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

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

\[x^{2} + 4 + \frac{4}{x^{2}} + x - \frac{2}{x} - 10 = 0\]

\[x^{2} + \frac{4}{x²} + x - \frac{2}{x} - 6 = 0\]

\[\left( x^{2} + \frac{4}{x^{2}} \right) + \left( x - \frac{2}{x} \right) - 6 = 0\]

\[Пусть\ x - \frac{2}{x} = t;\]

\[x^{2} + \frac{4}{x^{2}} = t^{2} + 4.\]

\[t^{2} + 4 + t - 6 = 0\]

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

\[t_{1} + t_{2} = - 1;\ \ t_{1} \cdot t_{2} = - 2\]

\[t_{1} = - 2;\ \ t_{2} = 1.\]

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

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

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

\[D_{1} = 1 + 2 = 3\]

\[x_{1} = - 1 + \sqrt{3};\ \ \ x_{2} = - 1 - \sqrt{3};\]

\[y_{1} = - \frac{2}{- 1 + \sqrt{3}} =\]

\[= - \frac{2 \cdot \left( \sqrt{3} + 1 \right)}{\left( \sqrt{3} - 1 \right)\left( \sqrt{3} + 1 \right)} =\]

\[= - \frac{2 \cdot \left( \sqrt{3} + 1 \right)}{2} = - 1 - \sqrt{3};\]

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

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

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

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

\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = 2;\ \ \ x_{2} = - 1;\]

\[y_{1} = - \frac{2}{x} = - 1;\ \ \ y_{2} = - \frac{2}{- 1} = 2.\]

\[Ответ:( - 1;2);(2; - 1);\]

\[\left( - 1 + \sqrt{3};\ - 1 - \sqrt{3} \right);\]

\[\left( - 1 - \sqrt{3};\ - 1 + \sqrt{3} \right).\]