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МатематикаГДЗ по алгебре 8 класс Макарычев Задание 543
Задание 543
\[\boxed{\text{543\ (543).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
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\[\textbf{а)}\ 25 = 26x - x^{2}\] \[x^{2} - 26x + 25 = 0\] \[D_{1} = 13^{2} - 1 \cdot 25 =\] \[= 169 - 25 = 144\] \[x_{1,2} = 13 \pm \sqrt{144} = 13 \pm 12\] \[x_{1} = 25;\ \ x_{2} = 1.\] |
\[\textbf{б)}\ 3x^{2} = 10 - 29x\] \[3x^{2} + 29x - 10 = 0\] \[D = 841 + 120 = 961\] \[x_{1,2} = \frac{- 29 \pm \sqrt{961}}{2 \cdot 3} = \frac{- 29 \pm 31}{6}\] \[x_{1} = \frac{1}{3};\ \ x_{2} = - 10.\] |
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\[\textbf{в)}\ y^{2} = 4y + 96\] \[y^{2} - 4y - 96 = 0\] \[D_{1} = 2^{2} + 96 = 100\] \[y_{1,2} = 2 \pm \sqrt{100} = 2 \pm 10\] \[y_{1} = 12;\ \ y_{2} = - 8.\] |
\[\textbf{г)}\ 3p^{2} + 3 = 10p\] \[3p^{2} - 10p + 3 = 0\] \[D_{1} = 5^{2} - 3 \cdot 3 = 25 - 9 = 16\] \[p_{1,2} = \frac{5 \pm \sqrt{16}}{3} = \frac{5 \pm 4}{3}\] \[p_{1} = 3;\ \ p_{2} = \frac{1}{3}.\] |
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\[\textbf{д)}\ x^{2} - 20x = 20x + 100\] \[x^{2} - 40x - 100 = 0\] \[D_{1} = 20^{2} + 100 = 500\] \[x_{1,2} = 20 \pm \sqrt{500} = 20 \pm 10\sqrt{5}\] \[x_{1} = 20 + 10\sqrt{5};\ \ \] \[x_{2} = 20 - 10\sqrt{5}.\] |
\[\textbf{е)}\ 25x^{2} - 13x = 10x^{2} - 7\] \[15x^{2} - 13x + 7 = 0\] \[D = 169 - 420 < 0\] \[корней\ нет.\ \ \ \] |
