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МатематикаГДЗ по алгебре 8 класс Макарычев Задание 544
Задание 544
\[\boxed{\text{544\ (544).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
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\[\textbf{а)}\ (2x - 3)(5x + 1) = 2x + \frac{2}{5}\] \[10x^{2} + 2x - 15x - 3 - 2x - \frac{2}{5} = 0\] \[10x^{2} - 15x - \frac{17}{5} = 0\] \[50x^{2} - 75x - 17 = 0\] \[D = 5625 + 3400 = 9025\] \[x_{1,2} = \frac{75 \pm \sqrt{9025}}{2 \cdot 50} = \frac{75 \pm 95}{100}\] \[x_{1} = 1,7;\ \ x_{2} = - 0,2.\] |
\[\textbf{б)}\ (3x - 1)(x + 3) = x(1 + 6x)\] \[3x^{2} + 9x - x - 3 = x + 6x^{2}\] \[3x^{2} - 7x + 3 = 0\] \[D = 49 - 36 = 13\] \[x_{1,2} = \frac{7 \pm \sqrt{13}}{3 \cdot 2}\] \[x_{1} = \frac{7 - \sqrt{13}}{6};\ \ x_{2} = \frac{7 + \sqrt{13}}{6}.\] |
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\[\textbf{в)}\ (x - 1)(x + 1) = 2\left( 5x - 10\frac{1}{2} \right)\] \[x^{2} + x - x - 1 = 10x - 2 \cdot \frac{21}{2}\] \[x^{2} - 10x - 1 + 21 = 0\] \[x^{2} - 10x + 20 = 0\] \[D_{1} = 5^{2} - 20 = 25 - 20 = 5\] \[x_{1,2} = 5 \pm \sqrt{5}\] \[x_{1} = 5 + \sqrt{5};\ \ x_{2} = 5 - \sqrt{5}.\] |
\[\textbf{г)} - x(x + 7) = (x - 2)(x + 2)\] \[- x^{2} - 7x = x^{2} - 4\] \[2x^{2} + 7x - 4 = 0\] \[D = 49 + 32 = 81\] \[x_{1,2} = \frac{7 \pm \sqrt{81}}{2 \cdot ( - 2)} = \frac{7 \pm 9}{- 4}\] \[x_{1} = - 4;\ \ x_{2} = 0,5.\ \ \] |
