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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 178
Задание 178
\[\boxed{\text{178\ (178).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
\[\mathbf{Числовые\ промежутки.}\]
Решение.
\[- 4 \leq x \leq 2\]
\[x \in \lbrack - 4;2\rbrack\]
\[Ответ:рисунок\ г).\]
\[\boxed{\text{178.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ \left\{ \begin{matrix} 3 \cdot (x + y)^{2} + 2 \cdot (x - 2y)^{2} = 5 \\ 2 \cdot (x - 2y) - x - y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \]
\[\left\{ \begin{matrix} 3 \cdot (x + y)^{2} + 2 \cdot (x - 2y)^{2} = 5 \\ 2 \cdot (x - 2y) - (x + y) = 1\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Пусть\ \left\{ \begin{matrix} x + y = a \\ x - 2y = b \\ \end{matrix} \right.\ ,\ тогда:\]
\[\left\{ \begin{matrix} 3a^{2} + 2b^{2} = 5 \\ 2b - a = 1\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} 3a^{2} + 2b^{2} = 5 \\ a = 2b - 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[3 \cdot (2b - 1)^{2} + 2b^{2} - 5 = 0\]
\[3 \cdot \left( 4b^{2} - 4b + 1 \right) + 2b^{2} - 5 = 0\]
\[12b^{2} - 12b + 3 + 2b^{2} - 5 = 0\]
\[14b^{2} - 12b - 2 = 0\ \ \ \ \ \ \ \ \ |\ :2\]
\[7b^{2} - 6b - 1 = 0\]
\[D = 36 + 28 = 64\]
\[b = \frac{6 + 8}{14} = 1\]
\[b = \frac{6 - 8}{14} = - \frac{1}{7}\]
\[\left\{ \begin{matrix} b = 1 \\ a = 1 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} b = - \frac{1}{7} \\ a = - \frac{9}{7} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x + y = 1\ \ \ \\ x - 2y = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 1 - y\ \ \ \ \ \ \ \ \ \\ 1 - y - 2y = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 1 - y \\ y = 0\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 1 \\ y = 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x + y = - \frac{9}{7} \\ x - 2y = - \frac{1}{7} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x = - \frac{9}{7} - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - \frac{9}{7} - y - 2y + \frac{1}{7} = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\text{\ \ \ }\left\{ \begin{matrix} x = - \frac{9}{7} - y \\ y = - \frac{8}{21}\text{\ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - \frac{19}{21} \\ y = - \frac{8}{21} \\ \end{matrix} \right.\ \]
\[Ответ:(1;0);\ \left( - \frac{19}{21};\ - \frac{8}{21} \right)\text{.\ }\]
\[2)\ \left\{ \begin{matrix} \frac{y}{x} + xy = - 10 \\ \frac{5y}{x} - 2xy = 13 \\ \end{matrix} \right.\ \]
\[Пусть\ \frac{y}{x} = a,\ а\ xy = b,\ то:\]
\[\left\{ \begin{matrix} a + b = - 10\ \ \ \ | \cdot 2 \\ 5a - 2b = 13\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} 2a + 2b = - 20 \\ 5a - 2b = 13 \\ \end{matrix} + \right.\ \]
\[\left\{ \begin{matrix} 7a = - 7 \\ 5a - 2b = 13 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} a = - 1 \\ b = - 9 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \frac{y}{x} = - 1 \\ xy = - 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = - x \\ - x^{2} = - 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} y = - x \\ x^{2} = 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = - x \\ x = \pm 3 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = - 3 \\ x = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 3 \\ x = - 3 \\ \end{matrix} \right.\ \]
\[Ответ:(3;\ - 3),\ ( - 3;3).\]
\[3)\ \left\{ \begin{matrix} x^{2}y^{2} + xy = 6 \\ 2x - y = 3\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} x^{2}y^{2} + xy = 6 \\ y = 2x - 3\ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Пусть\ xy = a,\ то:\]
\[a^{2} + a - 6 = 0\]
\[a_{1} + a_{2} = - 1,\ \ a_{1} = - 3\]
\[a_{1}a_{2} = - 6,\ \ a_{2} = 2\]
\[\left\{ \begin{matrix} xy = - 3\ \ \ \ \\ y = 2x - 3 \\ \end{matrix} \right.\ \text{\ \ }\]
\[2x^{2} - 3x + 3 = 0\]
\[D = 9 - 24 < 0\]
\[\left\{ \begin{matrix} xy = 2\ \ \ \ \ \ \ \\ y = 2x - 3 \\ \end{matrix} \right.\ \]
\[2x^{2} - 3x - 2 = 0\]
\[D = 9 + 16 = 25\]
\[x = \frac{3 + 5}{4} = 2\]
\[x = \frac{3 - 5}{4} = - \frac{1}{2}\]
\[\left\{ \begin{matrix} x = 2 \\ y = 1 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} x = - 0,5 \\ y = - 4\ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:(2;1);\ ( - 0,5;\ - 4).\]
