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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 535
Задание 535
\[\boxed{\text{535\ (535).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ 14x^{2} - 5x - 1 = 0\]
\[D = 25 + 4 \cdot 14 = 25 + 56 = 81\]
\[x_{1,2} = \frac{5 \pm \sqrt{81}}{2 \cdot 14} = \frac{5 \pm 9}{28}\]
\[x_{1} = \frac{14}{28} = \frac{1}{2} = 0,5;\ \ \ \ \ \]
\[x_{2} = - \frac{4}{28} = - \frac{1}{7}\]
\[Ответ:x = - \frac{1}{7};\ \ \ \ x = - 0,5.\]
\[\textbf{б)} - y^{2} + 3y + 5 = 0\]
\[D = 9 + 20 = 29\]
\[y_{1,2} = \frac{- 3 \pm \sqrt{29}}{2 \cdot ( - 1)}\]
\[y_{1} = \frac{3 + \sqrt{29}}{2};\ y_{2} = \frac{3 - \sqrt{29}}{2}\]
\[Ответ:y = \frac{3 \pm \sqrt{29}}{2}.\]
\[\textbf{в)}\ 2x^{2} + x + 67 = 0\]
\[D = 1 - 536 = - 535\]
\[D < 0 - корней\ нет.\]
\[Ответ:корней\ нет.\]
\[\textbf{г)}\ 1 - 18p + 81p^{2} = 0\]
\[D = 324 - 324 = 0\]
\[p_{1} = \frac{18}{2 \cdot 81} = \frac{9}{81} = \frac{1}{9}\]
\[Ответ:p = \frac{1}{9}.\]
\[\textbf{д)} - 11y + y^{2} - 152 = 0\]
\[y^{2} - 11y - 152 = 0\]
\[D = 121 + 608 = 729\]
\[y_{1,2} = \frac{11 \pm \sqrt{729}}{2} = \frac{11 \pm 27}{2}\]
\[y_{1} = - \frac{16}{2} = - 8;\ y_{2} = \frac{38}{2} = 19\]
\[Ответ:y = - 8;\ \ y = 19.\]
\[\textbf{е)}\ 18 + 3x^{2} - x = 0\]
\[3x^{2} - x + 18 = 0\]
\[D = 1 - 216 = - 215\]
\[D < 0 - корней\ нет.\ \ \ \ \]
\[Ответ:корней\ нет.\]
\[\boxed{\text{535.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
Пояснение.
Решение.
\[\textbf{а)}\ x^{2} - 11x + 31 = 1\]
\[x^{2} - 11x + 30 = 0\]
\[D = 121 - 120 = 1\]
\[x_{1,2} = \frac{11 \pm \sqrt{1}}{2} = \frac{11 \pm 1}{2}\]
\[x_{1} = 6;\ \ x_{2} = 5\]
\[Ответ:при\ x = 5;\ \ x = 6.\]
\[\textbf{б)}\ x^{2} - 5x - 3 = 2x - 5\]
\[x^{2} - 7x + 2 = 0\]
\[D = 49 - 8 = 41\]
\[x_{1,2} = \frac{7 \pm \sqrt{41}}{2}\]
\[x_{1} = \frac{7 - \sqrt{41}}{2};\ \ x_{2} = \frac{7 + \sqrt{41}}{2}\]
\[Ответ:при\ x = \frac{7 \pm \sqrt{41}}{2}.\]
\[\textbf{в)}\ 7x + 1 = 3x^{2} - 2x + 1\]
\[3x^{2} - 9x = 0\]
\[3x(x - 3) = 0\]
\[x_{1} = 0,\ \ \ \ x_{2} = 3\ \]
\[Ответ:при\ x = 0;\ \ x = 3.\]
\[\textbf{г)} - 2x^{2} + 5x + 6 = 4x^{2} + 5x\]
\[6x^{2} - 6 = 0\]
\[6 \cdot \left( x^{2} - 1 \right) = 0\]
\[x^{2} - 1 = 0\]
\[x^{2} = 1\]
\[x_{1,2} = \pm 1\ \ \]
\[Ответ:при\ x = \pm 1.\]
