Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1424

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\[1)\ \left\{ \begin{matrix} x^{2} - y^{2} = 13 \\ x - y = 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x^{2} - y^{2} - 13 = 0 \\ x = 1 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(1 + y)^{2} - y^{2} - 13 = 0\]

\[1 + 2y + y^{2} - y^{2} - 13 = 0\]

\[2y - 12 = 0\]

\[y - 6 = 0\]

\[y = 6;\]

\[x = 1 + 6 = 7.\]

\[Ответ:\ \ (7;\ 6).\]

\[2)\ \left\{ \begin{matrix} x^{2} - 3y = - 5 \\ 7x + 3y = 23\ \\ \end{matrix} \right.\ ( + )\ \ \ \]

\[\left\{ \begin{matrix} x^{2} - 3y = - 5 \\ y = \frac{23 - 7x}{3}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[x^{2} + 7x - 3y + 3y = - 5 + 23\]

\[x^{2} + 7x = 18\]

\[x^{2} + 7x - 18 = 0\]

\[D = 49 + 72 = 121\]

\[x_{1} = \frac{- 7 - 11}{2} = - 9;\]

\[x_{2} = \frac{- 7 + 11}{2} = 2;\]

\[y_{1} = \frac{23 - 7 \bullet ( - 9)}{3} = \frac{23 + 63}{3} =\]

\[= \frac{86}{3} = 28\frac{2}{3};\]

\[y_{2} = \frac{23 - 7 \bullet 2}{3} =\]

\[= \frac{23 - 14}{3} = \frac{9}{3} = 3.\]

\[Ответ:\ \ \left( - 9;\ 28\frac{2}{3} \right);\ \ (2;\ 3).\]