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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 581
Задание 581
\[\boxed{\text{581\ (581).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
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\[\textbf{а)}\ x^{2} - 2x - 9 = 0\] \[D_{1} = 1 + 9 = 10\] \[x_{1} = 1 + \sqrt{10};\] \[x_{2} = 1 - \sqrt{10}.\] \[1)\ x_{1} + x_{2} = 2:\] \[1 + \sqrt{10} + 1 - \sqrt{10} = 2\] \[2 = 2.\] \[2)\ x_{1}x_{2} = - 9:\] \[\left( 1 + \sqrt{10} \right)\left( 1 - \sqrt{10} \right) = - 9\] \[1^{2} - \left( \sqrt{10} \right)^{2} = - 9\] \[1 - 10 = - 9\] \[- 9 = - 9.\] |
\[\textbf{б)}\ 3x^{2} - 4x - 4 = 0\] \[D_{1} = 2^{2} + 3 \cdot 4 = 16\] \[x_{1} = \frac{2 + 4}{3} = 2;\ \ x_{2} = \frac{2 - 4}{3} = - \frac{2}{3}.\] \[1)\ x_{1} + x_{2} = \frac{4}{3}:\] \[2 + \left( - \frac{2}{3} \right) = \frac{4}{3}\] \[\frac{4}{3} = \frac{4}{3}.\] \[2)\ x_{1}x_{2} = - \frac{4}{3}:\] \[2 \cdot \left( - \frac{2}{3} \right) = - \frac{4}{3}\] \[- \frac{4}{3} = - \frac{4}{3}\text{.\ \ }\] |
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\[\textbf{в)}\ 2x^{2} + 7x - 6 = 0\] \[D = 49 + 48 = 97\] \[x_{1} = \frac{- 7 + \sqrt{97}}{4};\] \[x_{2} = \frac{- 7 - \sqrt{97}}{4}.\] \[1)\ x_{1} + x_{2} = - 3,5:\] \[\frac{- 7 + \sqrt{97}}{4} + \frac{- 7 - \sqrt{97}}{4} = - 3,5\ \ \ \ \ | \cdot 4\] \[- 7 + \sqrt{97} - 7 - \sqrt{97} = - 14\] \[- 14 = - 14.\] \[2)\ x_{1} \cdot x_{2} = - 3:\] \[\frac{- 7 + \sqrt{97}}{4} \cdot \frac{- 7 - \sqrt{97}}{4} = - 3\] \[\frac{( - 7)^{2} - {\sqrt{97}}^{2}}{16} = - 3\] \[\frac{49 - 97}{16} = - 3\] \[- \frac{48}{16} = - 3\] \[- 3 = - 3.\] |
\[\textbf{г)}\ 2x^{2} + 9x + 8 = 0\] \[D = 81 - 64 = 17\] \[x_{1} = \frac{- 9 + \sqrt{17}}{4};\] \[x_{2} = \frac{- 9 - \sqrt{17}}{4}.\] \[1)\ x_{1} + x_{2} = - 4,5:\] \[\frac{- 9 + \sqrt{17}}{4} + \frac{- 9 - \sqrt{17}}{4} = 4,5\ \ \ \ | \cdot 4\] \[- 9 + \sqrt{17} - 9 - \sqrt{17} = - 18\] \[- 18 = - 18.\] \[2)\ x_{1}x_{2} = 4:\] \[\frac{- 9 + \sqrt{17}}{4} \cdot \frac{- 9 - \sqrt{17}}{4} = 4\] \[\frac{( - 9)^{2} - {\sqrt{17}}^{2}}{16} = 4\] \[\frac{81 - 17}{16} = 4\] \[\frac{64}{16} = 4\] \[4 = 4.\ \ \] |
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\(\boxed{\text{581.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\)
Пояснение.
Решение.
\[\textbf{а)}\ x^{2} - 9x + 20 = 0\]
\[\left\{ \begin{matrix} x_{1} + x_{2} = 9 \\ x_{1}x_{2} = 20 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x_{1} = 4 \\ x_{2} = 5 \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ y^{2} + 11y - 12 = 0\]
\[\left\{ \begin{matrix} y_{1} + y_{2} = - 11 \\ y_{1}y_{2} = - 12\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y_{1} = - 12 \\ y_{2} = 1\ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\textbf{в)}\ y^{2} + y - 56 = 0\]
\[\left\{ \begin{matrix} y_{1} + y_{2} = - 1 \\ y_{1}y_{2} = - 56\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y_{1} = 7\ \ \\ y_{2} = - 8 \\ \end{matrix} \right.\ \]
\[\textbf{г)}\ z^{2} - 19z + 88 = 0\]
\[\left\{ \begin{matrix} z_{1} + z_{2} = 19 \\ z_{1}z_{2} = 88\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} z_{1} = 11 \\ z_{2} = 8\ \ \\ \end{matrix} \right.\ \]
