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МатематикаРешебник по алгебре 9 класс Макарычев Задание 532
Задание 532
\[\boxed{\text{532\ (532).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} (x + y)(x - y) = 0 \\ 2x - y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[1)\left\{ \begin{matrix} x + y = 0 \\ 2x - y = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - y\ \\ - 3y = 1 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = - \frac{1}{3} \\ x = \frac{1}{3}\text{\ \ \ \ } \\ \end{matrix} \right.\ ,\]
\[2)\ \left\{ \begin{matrix} x - y = 0\ \\ 2x - y = 1 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = y \\ y = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 1 \\ y = 1 \\ \end{matrix}, \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 100\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 7y)(x + 7y) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 7y \\ y² = 2\ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = \pm \sqrt{2\ } \\ x = \pm 7\sqrt{2} \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} x = - 7y\ \ \ \ \\ 50y^{2} = 100 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y^{2} = 2\ \ \ \ \\ x = - 7y \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = \pm \sqrt{2\ } \\ x = \pm 7\sqrt{2.} \\ \end{matrix} \right.\ \]
\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 25\ \ \ \ \ \ \ \ \ \ \\ (x - 3)(y - 5) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[1)\left\{ \begin{matrix} x = 3\ \ \\ y^{2} = 16 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = 3\ \ \ \\ y = \pm 4 \\ \end{matrix} \right.\ ,\ \]
\[2)\left\{ \begin{matrix} y = 5 \\ x = 0. \\ \end{matrix} \right.\ \]
\[\textbf{г)}\ \left\{ \begin{matrix} x^{2} - y^{2} = 50\ \\ x(y + 1) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[1)\left\{ \begin{matrix} x = 0\ \ \ \ \ \ \\ y^{2} = - 50 \\ \end{matrix} \right.\ \Longrightarrow корней\ нет;\]
\[2)\ \left\{ \begin{matrix} y = - 1 \\ x² = 51 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = - 1\ \ \ \ \ \ \ \\ x = \pm \sqrt{51}. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ \left( \frac{1}{3};\ - \frac{1}{3} \right);\ \ (1;1);\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ \left( 7\sqrt{2};\sqrt{2} \right);\ \ \]
\[\left( - 7\sqrt{2};\sqrt{2} \right);\left( 7\sqrt{2};\ - \sqrt{2} \right);\]
\[\left( - 7\sqrt{2};\ - \sqrt{2} \right);\]
\(в)\ (0;5);\ \ (3;4);(3;\ - 4);\)
\(г)\ \left( \sqrt{51};\ - 1 \right);\left( \sqrt{51};1 \right).\)
\[\boxed{\text{532.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[a_{n} = n^{2} - n - 20\]
\[n^{2} - n - 20 = 0\]
\[n_{1} + n_{2} = 1;\ \ \ n_{1} \cdot n_{2} = - 20.\]
\[n_{1} = 5,\ \ n_{2} = - 4;\]
\[\Longrightarrow (n - 5)(n + 4) < 0\ \ \ \]
\[при\ \ n \in ( - 4;5).\]
\[a_{1} = 1 - 1 - 20 = - 20\]
\[a_{2} = 4 - 2 - 20 = - 18\]
\[a_{3} = 9 - 3 - 20 = - 14\]
\[a_{4} = 16 - 4 - 20 = - 8.\]
