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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 542
Задание 542
\[\boxed{\text{542\ (542).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
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\[\textbf{а)}\ 5x^{2} = 9x + 2\] \[5x^{2} - 9x - 2 = 0\] \[D = 81 + 40 = 121\] \[x_{1,2} = \frac{9 \pm \sqrt{121}}{10} = \frac{9 \pm 11}{10}\] \[x_{1} = 2;\ \ x_{2} = - 0,2.\] |
\[\textbf{б)} - x^{2} = 5x - 14\] \[- x^{2} - 5x + 14 = 0\] \[x^{2} + 5x - 14 = 0\] \[D = 25 + 56 = 81\] \[x_{1,2} = \frac{- 5 \pm \sqrt{81}}{2 \cdot 1} = \frac{- 5 \pm 9}{2}\] \[x_{1} = - 7;\ \ x_{2} = 2.\] |
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\[\textbf{в)}\ 6x + 9 = x^{2}\] \[x^{2} - 6x - 9 = 0\] \[D_{1} = 3^{2} + 9 = 9 + 9 = 18\] \[x_{1,2} = 3 \pm \sqrt{18} = 3 \pm 3\sqrt{2}\] \[x_{1} = 3 + 3\sqrt{2};\ \ x_{2} = 3 - 3\sqrt{2}.\] |
\[\textbf{г)}\ z - 5 = z^{2} - 25\] \[z^{2} - z - 20 = 0\] \[D = 1 + 80 = 81\] \[z_{1,2} = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2}\] \[z_{1} = 5;\ \ z_{2} = - 4.\] |
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\[\textbf{д)}\ y^{2} = 52y - 576\] \[y^{2} - 52y + 576 = 0\] \[D_{1} = 26^{2} - 576 =\] \[= 676 - 576 = 100\] \[y = 26 \pm 10\] \[y_{1} = 26 - 10 = 16;\] \[y_{2} = 26 + 10 = 36.\] |
\[\textbf{е)}\ 15y^{2} - 30 = 22y + 7\] \[15y^{2} - 22y - 37 = 0\] \[D_{1} = 11^{2} + 37 \cdot 15 =\] \[= 121 + 555 = 676\] \[y_{1,2} = \frac{11 \pm \sqrt{676}}{15} = \frac{11 \pm 26}{15}\] \[y_{1} = \frac{37}{15} = 2\frac{7}{15};\ \ y_{2} = - 1.\] |
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\[\textbf{ж)}\ 25p^{2} = 10p - 1\] \[25p^{2} - 10p + 1 = 0\] \[D_{1} = 5^{2} - 1 \cdot 25 = 25 - 25 = 0\] \[p = \frac{5}{25} = \frac{1}{5} = 0,2.\] |
\[\textbf{з)}\ 299x^{2} + 100x = 500 - 101x^{2}\] \[400x^{2} + 100x - 500 = 0\ \ \ |\ :100\] \[4x^{2} + x - 5 = 0\] \[D = 1 + 80 = 81\] \[x_{1,2} = \frac{- 1 \pm \sqrt{81}}{8} = \frac{- 1 \pm 9}{8}\] \[x_{1} = 1;\ \ x_{2} = - \frac{5}{4} = - 1,25.\ \] |
\(\boxed{\text{542.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\)
Пояснение.
Раскроем скобки, перенесем все значения влево, изменив знаки на противоположные, приравняем к нулю, чтобы получилось уравнение вида
\[ax^{2} + bx + c = 0.\]
Формула:
\[(a - b)(a + b) = a^{2} - b^{2}.\]
Решение.
