ГДЗ по алгебре 8 класс Макарычев Задание 582

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Задание 582

Содержание

\[\boxed{\text{582\ (582).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ x^{2} - 15x - 16 = 0\]

\[D = 225 + 64 = 289\]

\[x_{1} = \frac{15 - 17}{2} = - 1;\ \ \]

\[x_{2} = \frac{15 + 17}{2} = 16.\]

\[1)\ x_{1} + x_{2} = 15\]

\[- 1 + 16 = 15\]

\[15 = 15.\]

\[2)\ x_{1}x_{2} = - 16\]

\[- 1 \cdot 16 = - 16\]

\[- 16 = - 16.\]

\[\textbf{б)}\ x^{2} - 6x - 11 = 0\]

\[D_{1} = 3^{2} + 11 = 20\]

\[x_{1} = 3 + \sqrt{20} = 3 + 2\sqrt{5};\ \ \]

\[x_{2} = 3 - \sqrt{20} = 3 - 2\sqrt{5}.\]

\[1)\ x_{1} + x_{2} = 6\]

\[3 + 2\sqrt{5} + 3 - 2\sqrt{5} = 6\]

\[6 = 6.\]

\[2)\ x_{1}x_{2} = - 11\]

\[\left( 3 + 2\sqrt{5} \right)\left( 3 - 2\sqrt{5} \right) = - 11\]

\[9 - 4 \cdot 5 = - 11\]

\[- 11 = - 11.\]

\[\textbf{в)}\ 12x^{2} - 4x - 1 = 0\ \ \ \ |\ :12\]

\[x^{2} - \frac{4x}{12} - \frac{1}{12} = 0\]

\[x^{2} - \frac{1}{3}x - \frac{1}{12} = 0\]

\[D = \frac{1}{9} + \frac{4}{12} = \frac{1}{9} + \frac{1}{3} = \frac{4}{9}\]

\[x_{1,2} = \frac{\frac{1}{3} \pm \sqrt{\frac{4}{9}}}{2} = \frac{\frac{1}{3} \pm \frac{2}{3}}{2} = \frac{1 \pm 2}{6}\]

\[x_{1} = \frac{1}{2};\ \ x_{2} = - \frac{1}{6}\]

\[1)\ x_{1} + x_{2} = \frac{1}{3}\]

\[\frac{1}{2} + \left( - \frac{1}{6} \right) = \frac{1}{3}\]

\[\frac{2}{6} = \frac{1}{3}\]

\[\frac{1}{3} = \frac{1}{3}.\]

\[2)\ x_{1}x_{2} = - \frac{1}{12}\]

\[\frac{1}{2} \cdot \left( - \frac{1}{6} \right) = - \frac{1}{12}\]

\[- \frac{1}{12} = - \frac{1}{12}.\]

\[\textbf{г)}\ x^{2} - 6 = 0\]

\[x^{2} = 6\]

\[x = \pm \sqrt{6}\]

\[1)\ x_{1} + x_{2} = 0\]

\[\sqrt{6} + \left( - \sqrt{6} \right) = 0\]

\[0 = 0.\]

\[2)\ x_{1}x_{2} = - 6\]

\[\sqrt{6} \cdot \left( - \sqrt{6} \right) = - 6\]

\[- 6 = - 6.\]

\[\textbf{д)}\ 5x^{2} - 18x = 0\]

\[x(5x - 18) = 0\]

\[x_{1} = 0;\ \ x_{2} = \frac{18}{5} = 3,6\]

\[1)\ x_{1} + x_{2} = 3,6\]

\[0 + 3,6 = 3,6\]

\[3,6 = 3,6.\]

\[2)\ x_{1}x_{2} = 0\]

\[0 \cdot 3,6 = 0\]

\[0 = 0.\]

\[\textbf{е)}\ 2x^{2} - 41 = 0\]

\[x^{2} = \frac{41}{2} = 20,5\]

\[x = \pm \sqrt{20,5}\]

\[1)\ x_{1} + x_{2} = 0\]

\[\sqrt{20,5} - \sqrt{20,5} = 0\]

\[0 = 0.\]

\[2)\ x_{1}x_{2} = - \frac{41}{2}\]

\[\sqrt{20,5} \cdot \left( - \sqrt{20,5} \right) = - \frac{41}{2}\]

\[- \frac{41}{2} = - \frac{41}{2}\text{.\ }\text{\ \ }\]

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