Решебник по алгебре 8 класс Макарычев ФГОС Задание 533

 

\[\boxed{\text{533\ (533).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 2x^{2} + 3x + 1 = 0\]

\[D = b^{2} - 4ac = 3^{2} - 4 \cdot 2 \cdot 1 =\]

\[= 9 - 8 = 1\]

\[D > 0 - 2\ корня.\]

\[\textbf{б)}\ 2x^{2} + x + 2 = 0\]

\[D = b^{2} - 4ac = 1^{2} - 4 \cdot 2 \cdot 2 =\]

\[= 1 - 16 = - 15\]

\[D < 0 - корней\ нет.\]

\[\textbf{в)}\ 9x^{2} + 6x + 1 = 0\]

\[D = b^{2} - 4ac = 6^{2} - 4 \cdot 9 \cdot 1 =\]

\[= 36 - 36 = 0\]

\[D = 0 - один\ корень.\]

\[\textbf{г)}\ x^{2} + 5x - 6 = 0\]

\[D = b^{2} - 4ac =\]

\[= 5^{2} - 4 \cdot 1 \cdot ( - 6) = 25 + 24 =\]

\[= 49\]

\[D > 0 - два\ корня.\ \]

\[\boxed{\text{533.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 14x^{2} - 5x - 1 = 0\]

\[D = 25 + 4 \cdot 14 = 25 + 56 = 81\]

\[x_{1,2} = \frac{5 \pm \sqrt{81}}{2 \cdot 14} = \frac{5 \pm 9}{28}\]

\[x_{1} = \frac{14}{28} = \frac{1}{2} = 0,5;\ \ \ \ \ \]

\[\ x_{2} = - \frac{4}{28} = - \frac{1}{7}\]

\[Ответ:x = - \frac{1}{7};\ \ \ \ x = - 0,5.\]

\[\textbf{б)} - y^{2} + 3y + 5 = 0\]

\[D = 9 + 20 = 29\]

\[y_{1,2} = \frac{- 3 \pm \sqrt{29}}{2 \cdot ( - 1)}\]

\[y_{1} = \frac{3 + \sqrt{29}}{2};\ y_{2} = \frac{3 - \sqrt{29}}{2}\]

\[Ответ:y = \frac{3 \pm \sqrt{29}}{2}.\]

\[\textbf{в)}\ 2x^{2} + x + 67 = 0\]

\[D = 1 - 536 = - 535\]

\[D < 0 - корней\ нет.\]

\[Ответ:корней\ нет.\]

\[\textbf{г)}\ 1 - 18p + 81p^{2} = 0\]

\[D = 324 - 324 = 0\]

\[p_{1} = \frac{18}{2 \cdot 81} = \frac{9}{81} = \frac{1}{9}\]

\[Ответ:p = \frac{1}{9}.\]

\[\textbf{д)} - 11y + y^{2} - 152 = 0\]

\[y^{2} - 11y - 152 = 0\]

\[D = 121 + 608 = 729\]

\[y_{1,2} = \frac{11 \pm \sqrt{729}}{2} = \frac{11 \pm 27}{2}\]

\[y_{1} = - \frac{16}{2} = - 8;\ y_{2} = \frac{38}{2} = 19\]

\[Ответ:y = - 8;\ \ y = 19.\]

\[\textbf{е)}\ 18 + 3x^{2} - x = 0\]

\[3x^{2} - x + 18 = 0\]

\[D = 1 - 216 = - 215\]

\[D < 0 - корней\ нет.\ \ \ \ \]

\[Ответ:корней\ нет.\]