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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 534
Задание 534
\[\boxed{\text{534\ (534).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ 3x^{2} - 7x + 4 = 0\]
\[D = b^{2} - 4ac =\]
\[= ( - 7)^{2} - 4 \cdot 3 \cdot 4 = 49 - 48 =\]
\[= 1\]
\[x_{1,2} = \frac{- b \pm \sqrt{D}}{2a} = \frac{- ( - 7) \pm \sqrt{1}}{2 \cdot 3}\]
\[x_{1} = \frac{8}{6} = \frac{4}{3} = 1\frac{1}{3};\ \ \ \ \ x_{2} = \frac{6}{6} = 1\]
\[Ответ:x = 1\frac{1}{3};\ \ \ x = 1.\ \]
\[\textbf{б)}\ 5x^{2} - 8x + 3 = 0\]
\[D = 64 - 4 \cdot 5 \cdot 3 = 64 - 60 = 4\]
\[x_{1,2} = \frac{8 \pm \sqrt{4}}{2 \cdot 5} = \frac{8 \pm 2}{10}\]
\[x_{1} = \frac{6}{10} = 0,6;\ \ \ \ \ \ x_{2} = \frac{10}{10} = 1\ \]
\[Ответ:x = 1;\ \ x = 0,6.\]
\[\textbf{в)}\ 3x^{2} - 13x + 14 = 0\]
\[D = 169 - 4 \cdot 3 \cdot 14 =\]
\[= 169 - 168 = 1\]
\[x_{1,2} = \frac{13 \pm \sqrt{1}}{2 \cdot 3} = \frac{13 \pm 1}{6}\]
\[x_{1} = \frac{12}{6} = 2;\ \ \ \ \ \]
\[x_{2} = \frac{14}{6} = \frac{7}{3} = 2\frac{1}{3}\ \]
\[Ответ:x = 2;\ \ x = 2\frac{1}{3}.\]
\[\textbf{г)}\ 2y^{2} - 9y + 10 = 0\]
\[D = 81 - 4 \cdot 2 \cdot 10 = 81 - 80 =\]
\[= 1\]
\[y_{1,2} = \frac{9 \pm \sqrt{1}}{2 \cdot 2} = \frac{9 \pm 1}{4}\]
\[y_{1} = \frac{8}{4} = 2;\ \ \ y_{2} = \frac{10}{4} = \frac{5}{2} = 2,5\ \]
\[Ответ:y = 2;\ \ \ y = 2,5.\]
\[\textbf{д)}\ 5y^{2} - 6y + 1 = 0\]
\[D = 36 - 4 \cdot 5 \cdot 1 = 36 - 20 =\]
\[= 16\]
\[y_{1,2} = \frac{6 \pm \sqrt{16}}{2 \cdot 5} = \frac{6 \pm 4}{10}\]
\[y_{1} = \frac{2}{10} = 0,2;\ \ \ \ \ y_{2} = \frac{10}{10} = 1\ \]
\[Ответ:y = 0,2;\ \ y = 1.\]
\[\textbf{е)}\ 4x^{2} + x - 33 = 0\]
\[D = 1 - 4 \cdot 4 \cdot ( - 33) =\]
\[= 1 + 528 = 529\]
\[x_{1,2} = \frac{- 1 \pm \sqrt{529}}{2 \cdot 4} = \frac{- 1 \pm 23}{8}\]
\[x_{1} = \frac{22}{8} = \frac{11}{4} = 2\frac{3}{4} = 2,75;\ \ \ \]
\[x_{2} = - \frac{24}{8} = - 3\ \]
\[Ответ:x = 2,75;\ \ \ x = - 3.\]
\[\textbf{ж)}\ y^{2} - 10y - 24 = 0\]
\[D = 100 - 4 \cdot ( - 24) =\]
\[= 100 + 96 = 196\]
\[y_{1,2} = \frac{10 \pm \sqrt{196}}{2} = \frac{10 \pm 14}{2}\]
\[y_{1} = \frac{24}{2} = 12;\ \ \ \ y_{2} = - \frac{4}{2} = - 2\ \]
\[Ответ:y = 12;\ \ y = - 2.\]
\[\textbf{з)}\ p^{2} + p - 90 = 0\]
\[D = 1 - 4 \cdot ( - 90) = 1 + 360 =\]
\[= 361\]
\[p_{1,2} = \frac{- 1 \pm \sqrt{361}}{2} = \frac{- 1 \pm 19}{2}\]
\[p_{1} = \frac{18}{2} = 9;\ \ \ \ \ p_{2} = - \frac{20}{2} = - 10\ \ \ \]
\[Ответ:p = 9;\ \ p = - 10.\]
\[\boxed{\text{534.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
Пояснение.
Решение.
\[\textbf{а)}\ 5x^{2} - 11x + 2 = 0\]
\[D = 121 - 40 = 81\]
\[x_{1,2} = \frac{11 \pm \sqrt{81}}{2 \cdot 5} = \frac{11 \pm 9}{10}\]
\[x_{1} = \frac{2}{10} = \frac{1}{5} = 0,2;\ x_{2} = \frac{20}{10} = 2\]
\[Ответ:x = 0,2;\ \ x = 2.\]
\[\textbf{б)}\ 2p^{2} + 7p - 30 = 0\]
\[D = 49 + 240 = 289\]
\[p_{1,2} = \frac{- 7 \pm \sqrt{289}}{2 \cdot 2} = \frac{- 7 \pm 17}{4}\]
\[p_{1} = \frac{10}{4} = 2,5;\ p_{2} = - \frac{24}{4} = - 6\]
\[Ответ:p = 2,5;\ \ \ p = - 6.\]
\[\textbf{в)}\ 9y^{2} - 30y + 25 = 0\]
\[D = 900 - 900 = 0\]
\[y = \frac{30}{2 \cdot 9} = \frac{15}{9} = \frac{5}{3} = 1\frac{2}{3}\]
\[Ответ:y = 1\frac{2}{3}.\]
\[\textbf{г)}\ 35x^{2} + 2x - 1 = 0\]
\[D = 4 + 140 = 144\]
\[x_{1,2} = \frac{- 2 \pm \sqrt{144}}{2 \cdot 35} = \frac{- 2 \pm 12}{70}\]
\[x_{1} = - \frac{14}{70} = - \frac{1}{5} = - 0,2;\ \ \ \ \]
\[\text{\ \ }x_{2} = \frac{10}{70} = \frac{1}{7}\]
\[Ответ:x = - 0,2;\ \ \ x = \frac{1}{7}.\]
\[\textbf{д)}\ 2y^{2} - y - 5 = 0\]
\[D = 1 + 40 = 41\]
\[y_{1,2} = \frac{1 \pm \sqrt{41}}{2 \cdot 2}\]
\[y_{1} = \frac{1 + \sqrt{41}}{4};\ \ y_{2} = \frac{1 - \sqrt{41}}{4}\ \]
\[Ответ:y = \frac{1 \pm \sqrt{41}}{4}.\]
\[\textbf{е)}\ 16x^{2} - 8x + 1 = 0\]
\[D = 64 - 64 = 0\]
\[x = \frac{8}{2 \cdot 16} = \frac{1}{4} = 0,25\ \ \]
\[Ответ:x = 0,25.\]
