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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 604
Задание 604
\[\boxed{\mathbf{604\ (604).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ y = \frac{3x^{2} - 10x + 3}{x - 3} - \frac{x^{2} - 1}{x + 1}\]
\[3x² - 10x + 3 = 0\]
\[D = 100 - 36 = 64\]
\[x_{1,2} = \frac{10 \pm 8}{6},\ \ x_{1} = 3,\ \ \]
\[x_{2} = \frac{1}{3}\]
\[\frac{3 \cdot (x - 3)\left( x - \frac{1}{3} \right)}{(x - 3)} - \frac{x^{2} - 1}{x + 1} =\]
\[= 3 \cdot \left( x - \frac{1}{3} \right) - \frac{(x - 1)(x + 1)}{(x + 1)} =\]
\[= 3 \cdot \left( x - \frac{1}{3} \right) - x + 1 =\]
\[= 3x - 1 - x + 1 = 2x\]
\[y = 2x,\ \ x \neq 3,\ \ x \neq - 1\]
| \[x\] | \[0\] | \[1\] | \[2\] |
|---|---|---|---|
| \[y\] | \[0\] | \[2\] | \[4\] |
\[2)\ y = \frac{5x^{2} + 4x - 1}{x + 1} - \frac{x^{2} - 3x}{x}\]
\[5x^{2} + 4x - 1 = 0\]
\[D = 16 + 20 = 36\]
\[x_{1,2} = \frac{- 4 \pm 6}{10},\ \ x_{1} = - 1,\]
\[\text{\ \ }x_{2} = \frac{1}{5}\]
\[\frac{5x^{2} + 4x - 1}{x + 1} - \frac{x^{2} - 3x}{x} =\]
\[= \frac{5 \cdot (x + 1)\left( x - \frac{1}{5} \right)}{(x + 1)} -\]
\[- \frac{x(x - 3)}{x} = 5x - 1 - x + 3 =\]
\[= 4x + 2\]
\[y = 4x + 2,\ \ x \neq 0,\ \]
\[\ x \neq - 1\]
| \[x\] | \[0\] | \[1\] | \[- 2\] |
|---|---|---|---|
| \[y\] | \[2\] | \[6\] | \[- 6\] |
\[\boxed{\mathbf{604.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ a_{1} = 6;\ \ a_{9} = 22\]
\[S_{12} = \frac{2a_{1} + 11d}{2} \cdot 12 =\]
\[= \left( 2a_{1} + 11d \right) \cdot 6\]
\[a_{9} = a_{1} + 8d = 22\ \ \]
\[8d = 22 - 6\]
\[8d = 16\ \]
\[d = 2\]
\[S_{12} = (2 \cdot 6 + 11 \cdot 2) \cdot 6 =\]
\[= (12 + 22) \cdot 6 = 34 \cdot 6 = 204.\]
\[Ответ:204.\]
\[2)\ a_{6} = 49;\ \ a_{20} = 7\]
\[\left\{ \begin{matrix} a_{6} = a_{1} + 5d = 49\ \\ a_{20} = a_{1} + 19d = 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} a_{1} + 5d = 49 \\ a_{1} + 19d = 7 \\ \end{matrix} \right.\ ( - )\text{\ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} - 14d = 42\ \ \ \ \\ a_{1} = 49 - 5d \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\left\{ \begin{matrix} d = - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = 49 + 5 \cdot 3 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} d = - 3\ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = 49 + 15 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} d = - 3\ \\ a_{1} = 64 \\ \end{matrix} \right.\ \]
\[S_{12} = \left( 2 \cdot 64 + 11 \cdot ( - 3) \right) \cdot 6 =\]
\[= (128 - 33) \cdot 6 = 570.\]
\[Ответ:570.\]
