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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 469
Задание 469
\[\boxed{\text{469\ (469).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\left\{ \begin{matrix} x^{2} + 10xy + 25y^{2} = 49 \\ x - 5y = - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} (x + 5y)^{2} = 49 \\ x = 5y - 3\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x + 5y = 7 \\ x = 5y - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }или\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]
\[\text{\ \ }\left\{ \begin{matrix} x + 5y = - 7 \\ x = 5y - 3\ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:(2;1);( - 5;\ - 0,4).\]
\[2)\ \left\{ \begin{matrix} x^{2} + 4xy + 4y^{2} = 4x + 2y \\ x + 2y = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\ \left\{ \begin{matrix} (x + 2y)^{2} = 4x + 2y \\ x = 4 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 4x + 2y = 16 \\ x = 4 - 2y\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 4(4 - 2y) + 2y - 16 = 0 \\ x = 4 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[16 - 8y + 2y - 16 = 0\]
\[- 6y = 0\]
\[\left\{ \begin{matrix} y = 0 \\ x = 4 \\ \end{matrix} \right.\ \]
\[Ответ:(4;0).\]
\[3)\ \left\{ \begin{matrix} x^{2} + y^{2} = 10 \\ xy = 3\ \ \ \ \ | \cdot 2 \\ \end{matrix} \right.\ \ \ \ + \ \ \]
\[+ \left\{ \begin{matrix} x^{2} + y^{2} + 2xy = 16 \\ xy = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} (x + y)^{2} = 16 \\ xy = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\left\{ \begin{matrix} x + y = 4 \\ xy = 3\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }или\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x + y = - 4 \\ xy = 3\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\boxed{\text{469.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ y = x \cdot |x|\]
\[y = x^{2},\ при\ \ x > 0\]
\[y = - x^{2},\ при\ \ x < 0\]
\[2)\ y = \frac{x}{|x|} \cdot (x^{2} - x - 6)\]
\[1.\ y = x^{2} - x - 6,\ при\ \ x > 0\]
\[2.\ y = - x^{2} + x + 6,\ при\ \ x < 0\]
\[1.\ y = x^{2} - x - 6\]
\[x_{0} = \frac{1}{2};\ \ y_{0} = \frac{1}{4} - \frac{1}{2} - 6 = - 6\frac{1}{4},\ \ \]
\[\left( \frac{1}{2};\ - 6\frac{1}{4} \right) - вершина\ параболы.\]
\[x^{2} - x - 6 = 0\]
\[x_{1} + x_{2} = 1,\ \ x_{1} = 3\]
\[x_{1}x_{2} = - 6,\ \ \]
\[x_{2} = - 2\ (не\ удовлетворяет),\ \ \]
\[(3;0)\]
\[x = 0,\ \ y = - 6,\ \ (0; - 6)\]
\[2\text{.\ }y = - x^{2} + x + 6\]
\[x_{0} = \frac{1}{2},\ \ \]
\[y_{0} = - \frac{1}{4} + \frac{1}{2} + 6 =\]
\[= 6\frac{1}{4},\ \ \ \left( \frac{1}{2};6\frac{1}{4} \right)\]
\[- x^{2} + x + 6 = 0\]
\[x_{1} + x_{2} = 1,\ \ x_{1} = - 2\]
\[x_{1}x_{2} = - 6,\ \ \]
\[x_{2} = 3\ (не\ удовлетворяет),\ \ \]
\[( - 2;0)\]
\[x = 0,\ \ y = 6,\ \ (0;6)\]
\[3)\ y = x^{2} - 4|x| + 3\]
\[x_{0} = 2,\ \ \]
\[y_{0} = 4 - 8 + 3 = - 1,\]
\[\ \ x > 0\]
\[x_{0} = - 2,\ \ y_{0} = - 1,\ \]
\[\ x < 0\]
\[x = 0,\ \ y = 3\]
\[x^{2} - 4x + 3 = 0\]
\[x_{1} + x_{2} = 4,\ \ x_{1} = 1\]
\[x_{1}x_{2} = 3,\ \ x_{2} = 3\ \]
\[x^{2} + 4x + 3 = 0\]
\[x_{1} + x_{2} = - 4,\ \ x_{1} = - 1\]
\[x_{1}x_{2} = 3,\ \ x_{2} = - 3\ \]
\[4)\ y = x^{2} + 3x \cdot \frac{|x - 3|}{x - 3} - 4\]
\[1.y = x^{2} + 3x - 4,\ \ x > 0\]
\[2.\ y{= x}^{2} - 3x - 4,\ \ x < 0\]
\[1.\ y = x^{2} + 3x - 4\]
\[x_{0} = - 1,5,\ \ \]
\[y_{0} = \frac{9}{4} - \frac{9}{2} - 4 = - 6\frac{1}{4},\ \ \]
\[\left( - 1,5;\ - 6\frac{1}{4} \right)\]
\[x = 0,\ \ y = - 4,\ \ \]
\[(0;\ - 4)\]
\[x^{2} + 3x - 4 = 0\]
\[x_{1} + x_{2} = - 3,\ \ \]
\[x_{1} = - 4\ (не\ удовлетворяет)\]
\[x_{1}x_{2} = - 4,\ \ x_{2} = 1,\ \ \]
\[(1;0)\ \]
\[2.\ y = x^{2} - 3x - 4\]
\[x_{0} = 1,5,\ \ \]
\[y_{0} = \frac{9}{4} - \frac{9}{2} - 4 = - 6\frac{1}{4}\]
\[x = 0,\ \ y = - 4,\ \ (0;\ - 4)\]
\[x^{2} - 3x - 4 = 0\]
\[x_{1} + x_{2} = 3,\ \ \]
\[x_{1} = 4\ \ (не\ удовлетворяет),\]
\[\ \ ( - 1;0)\]
\[x_{1}x_{2} = - 4,\ \ x_{2} = - 1\ \]
