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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 854
Задание 854
\[\boxed{\text{854\ (854).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[C_{x}^{4} = 14 \cdot C_{x - 2}^{2},\ \ x > 0;\]
\[\frac{x!}{(x - 4)!} = \frac{14 \cdot (x - 2)!}{(x - 2 - 3)!}\]
\[\frac{(x - 1) \cdot x}{(x - 4)} = \frac{14}{1}\]
\[14x - 56 = x^{2} - x\]
\[x^{2} - 15x + 56 = 0\]
\[D = 225 - 224 = 1\]
\[x_{1} = \frac{15 - 1}{2} = 7,\ \ \]
\[x_{2} = \frac{15 + 1}{2} = 8.\]
\[Ответ:7\ или\ 8.\]
\[\boxed{\text{854.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\left\{ \begin{matrix} x^{3} + x^{3}y^{3} + y^{3} = 12 \\ x + xy + y = 0\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x³ + y³ = 12 - x³y³ \\ xy = - x - y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} x^{3} + y^{3} = 12 + (x + y)^{3} \\ xy = - x - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2}y + xy^{2} + 4 = 0 \\ xy = - x - y\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} \text{xy}(x + y) = - 4 \\ xy = - x - y\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x²y² = 4\ \ \ \ \ \ \ \ \ \\ xy = - x - y \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} xy = 2\ \ \ \ \ \ \ \ \ \ \\ 2 = - x - y \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = \frac{2}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{2}{y} + y + 2 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y^{2} + 2y + 2 = 0 \\ x = \frac{2}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[y^{2} + 2y + 2 = 0\]
\[D = 4 - 8 = - 4 < 0 \Longrightarrow\]
\[\Longrightarrow нет\ корней.\]
\[2)\ \left\{ \begin{matrix} xy = - 2\ \ \ \ \ \ \ \ \ \ \ \\ x + y - 2 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - \frac{2}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ y - \frac{2}{y} - 2 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - \frac{2}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ y² - 2y - 2 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} - 2y - 2 = 0\]
\[D = 1 + 2 = 3\]
\[y_{1} = 1 - \sqrt{3},\ \ x_{1} = 1 + \sqrt{3},\]
\[y_{2} = 1 + \sqrt{3},\ \ x_{2} = 1 - \sqrt{3}.\]
\[Ответ:\left( 1 - \sqrt{3};1 + \sqrt{3} \right);\]
\[\left( 1 + \sqrt{3};1 - \sqrt{3} \right).\]
