Решебник по алгебре 9 класс Мерзляк Задание 166

\[\boxed{\text{166\ (166).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Если\ \text{m\ }и\ n -\]

\[последовательные\ \]

\[целые\ числа,\ то:\ \]

\[n = m + 1\ \ или\ m = n - 1.\]

\[Получаем:\]

\[1)\ mn = m \cdot (m + 1) > m \Longrightarrow\]

\[\Longrightarrow верно.\]

\[2)\ mn = n \cdot (n - 1) > n \Longrightarrow\]

\[\Longrightarrow верно\ при\ n > 1.\]

\[3)\ неверно,\ так\ как\ одно\ из\ \]

\[последовательных\ чисел\ \]

\[четное,\ а\ другое\ \]

\[нечетное:произведение\ \]

\[является\ нечетным\ числом.\]

\[4)\ верно,\ так\ как\ одно\ из\ \]

\[последовательных\ чисел\ \]

\[четное,\ а\ другое\ \]

\[нечетное:произведение\ \]

\[является\ нечетным\ числом.\]

\[\boxed{\text{166.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left\{ \begin{matrix} y - x = 2\ \ \ \ \ \ \\ x^{2} - 2xy = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

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

\[\left\{ \begin{matrix} y = 2 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4x - 2x^{2} - 3 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 2 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - x^{2} - 4x - 3 = 0 \\ \end{matrix} \right.\ \]

\[- x^{2} - 4x - 3 = 0\]

\[x_{1} + x_{2} = - 4,\ \ x_{1} = - 1\]

\[x_{1}x_{2} = 3,\ \ \ \ \ \ \ \ \ \ \ \ x_{2} = - 3\]

\[\left\{ \begin{matrix} y = 2 + x \\ x = - 1\ \ \ \ \ \\ x = - 3\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y = 1\ \ \ \\ x = - 1 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \]

\[\left\{ \begin{matrix} y = - 1 \\ x = - 3 \\ \end{matrix} \right.\ \]

\[Ответ:( - 1;1);\ \ ( - 3;\ - 1).\]

\[2)\ \left\{ \begin{matrix} y - 4x = 2\ \ \\ xy + 2x = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 2 + 4x\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ (2 + 4x)x + 2x = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[2x + 4x^{2} + 2x - 8 = 0\]

\[4x^{2} + 4x - 8 = 0\ \ \ |\ \ :4\]

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

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

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

\[y_{1} = 2 - 8 = - 6;\]

\[y_{2} = 2 + 4 = 6.\]

\[Ответ:( - 2;\ - 6);\ \ (1;6).\]

\[3)\ \left\{ \begin{matrix} xy = 15\ \ \ \ \\ 2x - y = 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} xy = 15\ \ \ \ \\ y = 2x - 7 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x(2x - 7) = 15 \\ y = 2x - 7\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2x^{2} - 7x - 15 = 0 \\ y = 2x - 7\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2x^{2} - 7x - 15 = 0\]

\[D = 169\]

\[x_{1} = \frac{7 + 13}{4} = 5\]

\[x_{2} = \frac{7 - 13}{4} = - 1,5\]

\[\left\{ \begin{matrix} y = 2x - 7 \\ x = 5\ \ \ \ \ \ \ \ \ \ \ \\ x = - 1,5\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 3 \\ x = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ }или\ \ \ \]

\[\left\{ \begin{matrix} y = - 10 \\ x = - 1,5 \\ \end{matrix} \right.\ \]

\[Ответ:(5;3);\ ( - 1,5;\ - 10).\]

\[4)\ \left\{ \begin{matrix} x - y = 6\ \ \ \ \ \ \\ x^{2} + y^{2} = 18 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = x - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + (x - 6)^{2} = 18 \\ \end{matrix} \right.\ \]

\[x^{2} + x^{2} - 12x + 36 - 18 = 0\]

\[2x^{2} - 12x + 18 = 0\ \ \ |\ :2\]

\[x^{2} - 6x + 9 = 0\]

\[(x - 3)^{2} = 0\]

\[x = 3;\]

\[y = x - 6 = 3 - 6 = - 3.\]

\[Ответ:\ \ (3;\ - 3).\]