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МатематикаГДЗ по алгебре 8 класс Макарычев Задание 546
Задание 546
\[\boxed{\text{546\ (546).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
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\[\textbf{а)}\ \frac{x^{2} - 1}{2} - 11x = 11\ \ \ \ \ | \cdot 2\] \[x^{2} - 1 - 22x = 22\] \[x^{2} - 22x - 23 = 0\] \[D_{1} = 121 + 23 = 144\] \[x_{1,2} = 11 \pm \sqrt{144} = 11 \pm 12.\] \[x_{1} = 23;\ \ x_{2} = - 1.\] |
\[\textbf{б)}\ \frac{x^{2} + x}{2} = \frac{8x - 7}{3}\ \ \ \ \ \ | \cdot 6\] \[3x^{2} + 3x = 16x - 14\] \[3x^{2} + 3x - 16x + 14 = 0\] \[3x^{2} - 13x + 14 = 0\] \[D = 169 - 168 = 1\] \[x_{1,2} = \frac{13 \pm \sqrt{1}}{2 \cdot 3} = \frac{13 \pm 1}{6}\] \[x_{1} = 2;\ \ x_{2} = \frac{14}{6} = 2\frac{1}{3}.\] |
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\[\textbf{в)}\ \frac{4x^{2} - 1}{3} = x(10x - 9)\ \ \ \ | \cdot 3\] \[4x^{2} - 1 = 30x^{2} - 27x\] \[- 26x^{2} + 27x - 1 = 0\] \[26x^{2} - 27x + 1 = 0\] \[D = 729 - 104 = 625\] \[x_{1,2} = \frac{- 27 \pm \sqrt{625}}{- 26 \cdot 2} = \frac{- 27 \pm 25}{- 52}\] \[x_{1} = \frac{1}{26};\ \ \ x_{2} = 1.\] |
\[\textbf{г)}\ \frac{3}{4}x^{2} - \frac{2}{5}x = \frac{4}{5}x^{2} + \frac{3}{4}\ \ \ \ \ \ | \cdot 20\] \[15x^{2} - 8x = 16x^{2} + 15\] \[15x^{2} - 8x - 16x^{2} - 15 = 0\] \[- x^{2} - 8x - 15 = 0\] \[x^{2} + 8x + 15 = 0\ \] \[D_{1} = 4^{2} - 15 = 16 - 15 = 1\] \[x_{1,2} = - 4 \pm \sqrt{1} = - 4 \pm 1\] \[x_{1} = - 5;\ \ x_{2} = - 3.\ \ \] |
