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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 536
Задание 536
\[\boxed{\text{536\ (536).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ 5x^{2} - 11x + 2 = 0\]
\[D = 121 - 40 = 81\]
\[x_{1,2} = \frac{11 \pm \sqrt{81}}{2 \cdot 5} = \frac{11 \pm 9}{10}\]
\[x_{1} = \frac{2}{10} = \frac{1}{5} = 0,2;\ x_{2} = \frac{20}{10} = 2\]
\[Ответ:x = 0,2;\ \ x = 2.\]
\[\textbf{б)}\ 2p^{2} + 7p - 30 = 0\]
\[D = 49 + 240 = 289\]
\[p_{1,2} = \frac{- 7 \pm \sqrt{289}}{2 \cdot 2} = \frac{- 7 \pm 17}{4}\]
\[p_{1} = \frac{10}{4} = 2,5;\ p_{2} = - \frac{24}{4} = - 6\]
\[Ответ:p = 2,5;\ \ \ p = - 6.\]
\[\textbf{в)}\ 9y^{2} - 30y + 25 = 0\]
\[D = 900 - 900 = 0\]
\[y = \frac{30}{2 \cdot 9} = \frac{15}{9} = \frac{5}{3} = 1\frac{2}{3}\]
\[Ответ:y = 1\frac{2}{3}.\]
\[\textbf{г)}\ 35x^{2} + 2x - 1 = 0\]
\[D = 4 + 140 = 144\]
\[x_{1,2} = \frac{- 2 \pm \sqrt{144}}{2 \cdot 35} = \frac{- 2 \pm 12}{70}\]
\[x_{1} = - \frac{14}{70} = - \frac{1}{5} = - 0,2;\ \ \ \ \ \ \]
\[x_{2} = \frac{10}{70} = \frac{1}{7}\]
\[Ответ:x = - 0,2;\ \ \ x = \frac{1}{7}.\]
\[\textbf{д)}\ 2y^{2} - y - 5 = 0\]
\[D = 1 + 40 = 41\]
\[y_{1,2} = \frac{1 \pm \sqrt{41}}{2 \cdot 2}\]
\[y_{1} = \frac{1 + \sqrt{41}}{4};\ \ y_{2} = \frac{1 - \sqrt{41}}{4}\ \]
\[Ответ:y = \frac{1 \pm \sqrt{41}}{4}.\]
\[\textbf{е)}\ 16x^{2} - 8x + 1 = 0\]
\[D = 64 - 64 = 0\]
\[x = \frac{8}{2 \cdot 16} = \frac{1}{4} = 0,25\ \ \]
\[Ответ:x = 0,25.\]
\[\boxed{\text{536.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
Пояснение.
Решение.
\[\textbf{а)}\ x^{2} - 6x = 5x - 18\]
\[x^{2} - 11x + 18 = 0\]
\[D = 121 - 72 = 49\]
\[x_{1,2} = \frac{11 \pm \sqrt{49}}{2} = \frac{11 \pm 7}{2}\]
\[x_{1} = \frac{4}{2} = 2;\ \ x_{2} = \frac{18}{2} = 9\]
\[Ответ:при\ x = 2;\ \ x = 9.\]
\[\textbf{б)}\ 3x^{2} - 4x + 3 = x^{2} + x + 1\]
\[2x^{2} - 5x + 2 = 0\]
\[D = 25 - 16 = 9\]
\[x_{1,2} = \frac{5 \pm \sqrt{9}}{2 \cdot 2} = \frac{5 \pm 3}{4}\]
\[x_{1} = \frac{2}{4} = \frac{1}{2} = 0,5;\ \ x_{2} = \frac{8}{4} = 2\ \ \ \]
\[Ответ:при\ x = 0,5;\ \ \ x = 2.\]
