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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 605
Задание 605
\[\boxed{\mathbf{605\ (605).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[x^{4} - y^{4} = x^{3} + y^{3}\]
\[\left( x^{2} - y^{2} \right)\left( x^{2} + y^{2} \right) =\]
\[= (x + y)\left( x^{2} - xy + y^{2} \right)\]
\[(x - y)(x + y)\left( x^{2} + y^{2} \right) -\]
\[- (x + y)\left( x^{2} - xy + y^{2} \right) = 0\]
\[(x + y)((x - y) \cdot\]
\[\cdot \left( x^{2} + y^{2} - \left( x^{2} - xy + y^{2} \right) \right) = 0\]
\[x = - y,\ \ то\ есть\ \text{\ x\ }и\ y\]
\[\ противоположные\ числа.\]
\[(x - y)\left( x^{2} + y^{2} \right) =\]
\[= \left( x^{2} - xy + y^{2} \right) \cdot 1\]
\[1)\ \left\{ \begin{matrix} x - y = x^{2} - xy + y^{2} \\ x^{2} + y^{2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x - y = 1 - xy \\ x^{2} + y^{2} = 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x + xy - y - 1 = 0 \\ x² + y² = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x(1 + y) - (y + 1) = 0 \\ x^{2} + y^{2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} (y + 1)(x - 1) = 0 \\ x^{2} + y^{2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\ \left\{ \begin{matrix} y = - 1 \\ x = 0\ \ \ \ \\ \end{matrix} \right.\ ;\ \ \ \ \left\{ \begin{matrix} y = 0 \\ x = 1 \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} x - y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + y^{2} = x^{2} - xy + y^{2} \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\ \left\{ \begin{matrix} x = 1 + y \\ xy = 0\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} y = - 1 \\ x = 0\ \ \ \\ \end{matrix} \right.\ ;\ \ \left\{ \begin{matrix} y = 0 \\ x = 1 \\ \end{matrix} \right.\ \]
\[Общие\ решения\ систем\ \ \ \]
\[(0;\ - 1);\ (1;0).\]
\[Ответ:таких\ \text{\ x\ }\ и\ y\ не\]
\[\ существует.\]
\[\boxed{\mathbf{605.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[x_{8} = - 14;\ \ x_{30} = - 3\]
\[\left\{ \begin{matrix} x_{1} + 7d = - 14 \\ x_{1} + 29d = - 3 \\ \end{matrix} \right.\ ( - )\text{\ \ \ }\]
\[\ \left\{ \begin{matrix} - 22d = - 11\ \ \ \\ x_{1} = - 14 - 7d \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} d = 0,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x_{1} = - 14 - 7 \cdot 0,5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} d = 0,5\ \ \ \ \ \ \\ x_{1} = - 17,5 \\ \end{matrix} \right.\ \]
\[S_{40} = \frac{2x_{1} + 39d}{2} \cdot 40 =\]
\[= \left( 2x_{1} + 39d \right) \cdot 20 =\]
\[= \left( 2 \cdot ( - 17,5) + 39 \cdot 0,5 \right) \cdot 20 =\]
\[= \left( - 35 + \frac{39}{2} \right) \cdot 20 =\]
\[= - 700 + 390 = - 310\]
\[Ответ:\ - 310.\]
