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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 167
Задание 167
\[\boxed{\text{167\ (167).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ 3\sqrt{98}\ \ \ и\ \ 4\sqrt{72}\]
\[3 \cdot 7\sqrt{2}\ \ и\ \ 4 \cdot 6\sqrt{2}\]
\[21\sqrt{2} < 24\sqrt{2}\text{\ \ }\]
\[3\sqrt{98} < 4\sqrt{72}\]
\[2)\frac{1}{2}\sqrt{68}\ \ \ и\ \frac{4}{3}\sqrt{45}\]
\[\sqrt{17}\ \ \ и\ \ 4\sqrt{5}\]
\[\sqrt{17} < \sqrt{80}\text{\ \ }\]
\[\frac{1}{2}\sqrt{68} < \frac{4}{3}\sqrt{45}\]
\[3)\frac{1}{6}\sqrt{108}\text{\ \ \ }и\ \ 6\sqrt{\frac{1}{12}}\]
\[\sqrt{3}\ \ и\ \ \sqrt{\frac{36}{12}}\]
\[\sqrt{3} = \sqrt{3}\text{\ \ }\]
\[\frac{1}{6}\sqrt{108} = 6\sqrt{\frac{1}{12}}\]
\[\boxed{\text{167.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ \left\{ \begin{matrix} x - y = 3 \\ xy = 28\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 3 + y\ \ \ \ \ \ \ \ \\ (3 + y)y = 28 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 3 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3y + y^{2} - 28 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} + 3y - 28 = 0\]
\[y_{1} + y_{2} = - 3,\ \ y_{1} = - 7\]
\[y_{1}y_{2} = - 28,\ \ y_{2} = 4\]
\[\left\{ \begin{matrix} x = - 4 \\ y = - 7 \\ \end{matrix} \right.\ \ \ \ или\ \ \left\{ \begin{matrix} x = 7 \\ y = 4 \\ \end{matrix} \right.\ \]
\[Ответ:( - 4;\ - 7);\ (7;4).\]
\[2)\ \left\{ \begin{matrix} y - 2x^{2} = 2 \\ 3x + y = 1\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = 2 + 2x^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 3x + 2 + 2x^{2} - 1 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = 2 + 2x^{2}\text{\ \ \ \ \ \ \ \ \ } \\ 2x^{2} + 3x + 1 = 0 \\ \end{matrix} \right.\ \]
\[2x^{2} + 3x + 1 = 0\]
\[D = 1\]
\[x_{1} = \frac{- 3 + 1}{4} = - 0,5\]
\[x_{2} = \frac{- 3 - 1}{4} = - 1\]
\[\left\{ \begin{matrix} x = 2,5 \\ y = - 0,5 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \left\{ \begin{matrix} y = 4 \\ x = - 1 \\ \end{matrix} \right.\ \]
\[Ответ:( - 0,5;2,5);\ ( - 1;4).\]
\[3)\ \left\{ \begin{matrix} x^{2} - 2y^{2} = 8 \\ x + y = 6\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x^{2} - 2y^{2} = 8 \\ x = 6 - y\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} (6 - y)^{2} - 2y^{2} - 8 = 0 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 36 - 12y + y^{2} - 2y^{2} - 8 = 0 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[- y^{2} - 12y + 28 = 0\]
\[y_{1} + y_{2} = - 12,\ \ y_{1} = 2\]
\[y_{1}y_{2} = - 28,\ \ y_{2} = - 14\]
\[\left\{ \begin{matrix} x = 4 \\ y = 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \left\{ \begin{matrix} x = 20\ \ \\ y = - 14 \\ \end{matrix} \right.\ \]
\[Ответ:(4;2);\ (20;\ - 14).\]
