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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 172
Задание 172
\[\boxed{\text{172\ (172).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
\[\mathbf{Числовые\ промежутки.}\]
Решение.
\[1)\ ( - 3;4)\]
\[2)\ \lbrack - 3;\ 4\rbrack\]
\[3)\ \lbrack - 3;4)\]
\[4)\ ( - 3;4\rbrack\]
\[\boxed{\text{172.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\left\{ \begin{matrix} 3x - y = 1\ \ \ \ \ \ \ \ \ \ \ \ \\ y = 3x^{2} + 8x - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = 3x - 1\ \ \ \ \ \ \ \ \ \ \ \ \\ y = 3x^{2} + 8x - 3 \\ \end{matrix} \right.\ \]
\[3x - 1 = 3x^{2} + 8x - 3\]
\[- 3x^{2} - 5x + 2 = 0\]
\[D = 25 + 24 = 49\]
\[x_{1} = \frac{5 + 7}{- 6} = - 2,\ \ \]
\[x_{2} = \frac{5 - 7}{- 6} = \frac{1}{3}\]
\[\left\{ \begin{matrix} y = - 7 \\ x = - 2 \\ \end{matrix}\text{\ \ } \right.\ \text{\ \ }или\ \ \ \left\{ \begin{matrix} y = 0 \\ x = \frac{1}{3} \\ \end{matrix} \right.\ \]
\[Ответ:( - 2;\ - 7);\ \left( \frac{1}{3};0 \right).\]
\[2)\ \left\{ \begin{matrix} 2x - y = 2 \\ y = \frac{4}{x}\text{\ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 2x - 2 \\ y = \frac{4}{x}\text{\ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[2x - 2 = \frac{4}{x}\ \ | \cdot x\]
\[2x^{2} - 2x - 4 = 0\ \ \ \ |\ :2\]
\[x^{2} - x - 2 = 0\]
\[x_{1} + x_{2} = 1,\ \ x_{1} = 2\]
\[x_{1}x_{2} = - 2,\ \ \ \ \ \ \ \ \ \ \ \ x_{2} = - 1\]
\[\left\{ \begin{matrix} x = 2 \\ y = 2 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x = - 1 \\ y = - 4 \\ \end{matrix} \right.\ \]
\[Ответ:(2;2);\ ( - 1; - 4).\]
\[3)\ \left\{ \begin{matrix} x + y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 1)^{2} + (y + 4)^{2} = 16 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} y = 1 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 1)^{2} + (1 - x + 4)^{2} = 16 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = 1 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2x + 1 + 25 - 10x + x^{2} - 16 = 0 \\ \end{matrix} \right.\ \]
\[2x^{2} - 12x + 10 = 0\ \ |\ :2\]
\[x^{2} - 6x + 5 = 0\]
\[x_{1} + x_{2} = 5,\ \ x_{1} = 5\]
\[x_{1}x_{2} = 6,\ \ \ \ \ \ \ \ \ \ \ \ x_{2} = 1\]
\[\left\{ \begin{matrix} y = - 4 \\ x = 5\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \left\{ \begin{matrix} y = 0 \\ x = 1 \\ \end{matrix} \right.\ \]
\[Ответ:(5;\ - 4);\ (1;0).\]
