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МатематикаГДЗ по алгебре 8 класс Макарычев Задание 552
Задание 552
\[\boxed{\text{552\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
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\[\textbf{а)}\ \frac{1}{7}x^{2} = 2x - 7\ \ \ \ \ | \cdot 7\] \[x^{2} - 14x + 49 = 0\] \[(x - 7)^{2} = 0\] \[x - 7 = 0\] \[x = 7.\] |
\[\textbf{в)}\ 4x^{2} = 7x + 7,5\ \ \ \ \ | \cdot 10\] \[40x^{2} - 70x - 75 = 0\ \ \ |\ :5\] \[8x^{2} - 14x - 15 = 0\] \[D = 196 + 480 = 676\] \[x_{1,2} = \frac{14 \pm \sqrt{676}}{2 \cdot 8} = \frac{14 \pm 26}{16}\] \[x_{1} = - \frac{12}{16} = - \frac{3}{4} = - 0,75;\] \[x_{2} = \frac{40}{16} = \frac{10}{4} = \frac{5}{2} = 2,5.\] |
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\[\textbf{б)}\ x^{2} + 1,2 = 2,6x\ \ \ \ \ \ \ | \cdot 10\] \[10x^{2} - 26x + 12 = 0\ \ \ \ \ |\ :2\] \[5x^{2} - 13x + 6 = 0\] \[D = 169 - 120 = 49\] \[x_{1,2} = \frac{13 \pm \sqrt{49}}{2 \cdot 5} = \frac{13 \pm 7}{10}\] \[x_{1} = \frac{20}{10} = 2;\] \[x_{2} = \frac{6}{10} = 0,6.\] |
\[\boxed{\text{552\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{1}{7}x^{2} = 2x - 7\ \ \ \ \ \ | \cdot 7\]
\[x^{2} - 14x + 49 = 0\]
\[(x - 7)^{2} = 0\]
\[x - 7 = 0\]
\[x = 7.\]
\[Ответ:x = 7.\]
\[\textbf{б)}\ x^{2} + \frac{6}{5} = 2,6x\ \ \ \ \ \ \ \ \ | \cdot 5\]
\[5x^{2} - 13x + 6 = 0\]
\[D = 169 - 120 = 49\]
\[x_{1} = \frac{13 + 7}{10} = 2;\ \ \ \ \ \]
\[x_{2} = \frac{13 - 7}{10} = \frac{6}{10} = 0,6.\]
\[Ответ:x = 2;\ \ x = 0,6.\]
\[\textbf{в)}\ 4x^{2} = 7x + 7,5\ \ \ \ \ \ \ \ | \cdot 2\]
\[8x^{2} - 14x - 15 = 0\]
\[D_{1} = 7^{2} + 8 \cdot 15 = 49 + 120 =\]
\[= 169\]
\[x_{1} = \frac{7 + 13}{8} = \frac{20}{8} = 2,5;\ \ \ \ \ \]
\[x_{2} = \frac{7 - 13}{8} = - \frac{6}{8} = - \frac{3}{4} =\]
\[= - 0,75.\]
\[Ответ:x = 2,5;\ \ \ x = - 0,75.\]
\[\textbf{г)}\ 6x^{2} - 2 = x\ \]
\[6x^{2} - x - 2 = 0\]
\[D = 1 + 48 = 49\]
\[x_{1} = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3};\ \ \ \ \]
\[x_{2} = \frac{1 - 7}{12} = - \frac{6}{12} = - \frac{1}{2} = - 0,5.\]
\[Ответ:x = \frac{2}{3};\ \ x = - 0,5.\]
