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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 582
Задание 582
\[\boxed{\text{582\ (582).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
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\[\textbf{а)}\ x^{2} - 15x - 16 = 0\] \[D = 225 + 64 = 289\] \[x_{1} = \frac{15 - 17}{2} = - 1;\ \ \] \[x_{2} = \frac{15 + 17}{2} = 16.\] \[1)\ x_{1} + x_{2} = 15\] \[- 1 + 16 = 15\] \[15 = 15.\] \[2)\ x_{1}x_{2} = - 16\] \[- 1 \cdot 16 = - 16\] \[- 16 = - 16.\] |
\[\textbf{б)}\ x^{2} - 6x - 11 = 0\] \[D_{1} = 3^{2} + 11 = 20\] \[x_{1} = 3 + \sqrt{20} = 3 + 2\sqrt{5};\ \ \] \[x_{2} = 3 - \sqrt{20} = 3 - 2\sqrt{5}.\] \[1)\ x_{1} + x_{2} = 6\] \[3 + 2\sqrt{5} + 3 - 2\sqrt{5} = 6\] \[6 = 6.\] \[2)\ x_{1}x_{2} = - 11\] \[\left( 3 + 2\sqrt{5} \right)\left( 3 - 2\sqrt{5} \right) = - 11\] \[9 - 4 \cdot 5 = - 11\] \[- 11 = - 11.\] |
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\[\textbf{в)}\ 12x^{2} - 4x - 1 = 0\ \ \ \ |\ :12\] \[x^{2} - \frac{4x}{12} - \frac{1}{12} = 0\] \[x^{2} - \frac{1}{3}x - \frac{1}{12} = 0\] \[D = \frac{1}{9} + \frac{4}{12} = \frac{1}{9} + \frac{1}{3} = \frac{4}{9}\] \[x_{1,2} = \frac{\frac{1}{3} \pm \sqrt{\frac{4}{9}}}{2} = \frac{\frac{1}{3} \pm \frac{2}{3}}{2} = \frac{1 \pm 2}{6}\] \[x_{1} = \frac{1}{2};\ \ x_{2} = - \frac{1}{6}\] \[1)\ x_{1} + x_{2} = \frac{1}{3}\] \[\frac{1}{2} + \left( - \frac{1}{6} \right) = \frac{1}{3}\] \[\frac{2}{6} = \frac{1}{3}\] \[\frac{1}{3} = \frac{1}{3}.\] \[2)\ x_{1}x_{2} = - \frac{1}{12}\] \[\frac{1}{2} \cdot \left( - \frac{1}{6} \right) = - \frac{1}{12}\] \[- \frac{1}{12} = - \frac{1}{12}.\] |
\[\textbf{г)}\ x^{2} - 6 = 0\] \[x^{2} = 6\] \[x = \pm \sqrt{6}\] \[1)\ x_{1} + x_{2} = 0\] \[\sqrt{6} + \left( - \sqrt{6} \right) = 0\] \[0 = 0.\] \[2)\ x_{1}x_{2} = - 6\] \[\sqrt{6} \cdot \left( - \sqrt{6} \right) = - 6\] \[- 6 = - 6.\] |
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\[\textbf{д)}\ 5x^{2} - 18x = 0\] \[x(5x - 18) = 0\] \[x_{1} = 0;\ \ x_{2} = \frac{18}{5} = 3,6\] \[1)\ x_{1} + x_{2} = 3,6\] \[0 + 3,6 = 3,6\] \[3,6 = 3,6.\] \[2)\ x_{1}x_{2} = 0\] \[0 \cdot 3,6 = 0\] \[0 = 0.\] |
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\[\textbf{е)}\ 2x^{2} - 41 = 0\] \[x^{2} = \frac{41}{2} = 20,5\] \[x = \pm \sqrt{20,5}\] \[1)\ x_{1} + x_{2} = 0\] \[\sqrt{20,5} - \sqrt{20,5} = 0\] \[0 = 0.\] \[2)\ x_{1}x_{2} = - \frac{41}{2}\] \[\sqrt{20,5} \cdot \left( - \sqrt{20,5} \right) = - \frac{41}{2}\] \[- \frac{41}{2} = - \frac{41}{2}\text{.\ }\text{\ \ }\] |
\(\boxed{\text{582.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\)
Пояснение.
Решение.
\[\textbf{а)}\ x^{2} + 16x + 63 = 0\]
\[\left\{ \begin{matrix} x_{1} + x_{2} = - 16 \\ x_{1}x_{2} = 63\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x_{1} = - 9 \\ x_{2} = - 7 \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ z^{2} + 2z - 48 = 0\]
\[\left\{ \begin{matrix} z_{1} + z_{2} = - 2 \\ z_{1}z_{2} = - 48\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} z_{1} = - 8 \\ z_{2} = 6\ \ \ \\ \end{matrix} \right.\ \]
