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МатематикаГДЗ по алгебре 8 класс Макарычев ФГОС Задание 764
Задание 764
\[\boxed{\text{764.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
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\[\textbf{а)}\ x^{2} - 5\sqrt{2}x + 12 = 0\] \[D = 50 - 48 = 2\] \[x_{1,2} = \frac{5\sqrt{2} \pm \sqrt{2}}{2}\] \[x_{1} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}\] \[x_{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}\] |
\[Проверка:\] \[\left\{ \begin{matrix} x_{1} + x_{2} = - b \\ x_{1}x_{2} = c\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \] \[\left\{ \begin{matrix} 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2} = - \left( - 5\sqrt{2} \right) \\ 2\sqrt{2} \cdot 3\sqrt{2} = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \] |
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\[\textbf{б)}\ x^{2} + 2\sqrt{3} - 72 = 0\] \[D = 12 + 288 = 300\] \[x_{1,2} = \frac{- 2\sqrt{3} \pm \sqrt{300}}{2}\] \[x_{1} = \frac{- 2\sqrt{3} + 10\sqrt{3}}{2} = 4\sqrt{3}\] \[x_{2} = \frac{- 2\sqrt{3} - 10\sqrt{3}}{2} = - 6\sqrt{3}\] |
\[Проверка:\] \[\left\{ \begin{matrix} x_{1} + x_{2} = - b \\ x_{1}x_{2} = c\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \] \[\left\{ \begin{matrix} 4\sqrt{3} + \left( - 6\sqrt{3} \right) = - 2\sqrt{3} \\ 4\sqrt{3} \cdot \left( - 6\sqrt{3} \right) = - 72\ \ \ \ \\ \end{matrix} \right.\ \] |
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\[\textbf{в)}\ y^{2} - 6y + 7 = 0\] \[D = 36 - 28 = 8\] \[y_{1,2} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2}\] \[y_{1} = 3 + \sqrt{2}\] \[y_{2} = 3 - \sqrt{2}\] |
\[Проверка:\] \[\left\{ \begin{matrix} x_{1} + x_{2} = - b \\ x_{1}x_{2} = c\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \] \[\left\{ \begin{matrix} 3 + \sqrt{2} + 3 - \sqrt{2} = 6 = - ( - 6) \\ \left( 3 + \sqrt{2} \right)\left( 3 - \sqrt{2} \right) = 9 - 2 = 7 \\ \end{matrix} \right.\ \] |
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\[\textbf{г)}\ p^{2} - 10p + 7 = 0\] \[D = 100 - 28 = 72\] \[p_{1,2} = \frac{10 \pm \sqrt{72}}{2} = \frac{10 \pm 2\sqrt{18}}{2}\] \[p_{1} = 5 + \sqrt{18}\] \[p_{2} = 5 - \sqrt{18}\] |
\[Проверка:\] \[\left\{ \begin{matrix} x_{1} + x_{2} = - b \\ x_{1}x_{2} = c\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \] \[\left\{ \begin{matrix} 5 + \sqrt{18} + 5 - \sqrt{18} = 10 = - ( - 10) \\ \left( 5 + \sqrt{18} \right)\left( 5 - \sqrt{18} \right) = 25 - 18 = 7 \\ \end{matrix} \right.\ \text{\ \ }\] |
