Решебник по алгебре 8 класс Мерзляк Задание 670

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\[1)\ 2x² + x\sqrt{5} - 15 = 0\]

\[D = \left( \sqrt{5} \right)^{2} + 4 \cdot 2 \cdot 15 =\]

\[= 5 + 120 = 125\]

\[x = \frac{- \sqrt{5} \pm \sqrt{125}}{4} = \frac{- \sqrt{5} \pm 5\sqrt{5}}{4}\]

\[x_{1} = \frac{4\sqrt{5}}{4} = \sqrt{5}\]

\[x_{2} = \frac{- 6\sqrt{5}}{4} = \frac{- 3\sqrt{5}}{2} = - 1,5\sqrt{5}\]

\[Ответ:x = \sqrt{5};\ x = - 1,5\sqrt{5}.\]

\[2)\ x² - x\left( \sqrt{6} - 1 \right) - \sqrt{6} = 0\ \]

\[D = \left( \sqrt{6} - 1 \right)^{2} + 4\sqrt{6} =\]

\[= 6 - 2\sqrt{6} + 1 + 4\sqrt{6} =\]

\[= 2\sqrt{6} + 7 = \left( \sqrt{6} + 1 \right)^{2}\]

\[x = \frac{\left( \sqrt{6} - 1 \right) \pm \sqrt{\left( \sqrt{6} + 1 \right)^{2}}}{2} =\]

\[= \frac{\left( \sqrt{6} - 1 \right) \pm (\sqrt{6} + 1)}{2}\ \]

\[x_{1} = \sqrt{6},\ \ x_{2} = - 1\]

\[Ответ:\ x = - 1;x = \sqrt{6}.\]

\[3)\ \frac{x^{2} - 4}{8} - \frac{2x + 3}{3} = - 1\]

\[\frac{x^{2} - 4}{8} - \frac{2x + 3}{3} + 1 = 0\]

\[3x^{2} - 16x - 12 = 0\]

\[D = 256 + 4 \cdot 3 \cdot 12 = 400\]

\[x = \frac{16 \pm \sqrt{400}}{6} = \frac{16 \pm 20}{6}\]

\[x_{1} = 6,\ \ x_{2} = - \frac{2}{3}\]

\[Ответ:\ x = - \frac{2}{3};x = 6.\]

\[4)\ \frac{4x^{2} + x}{3} - \frac{x^{2} + 17}{9} = \frac{5x - 1}{6}\]

\[\frac{4x^{2} + x}{3} - \frac{x^{2} + 17}{9} - \frac{5x - 1}{6} = 0\]

\[\frac{22x^{2} - 9x - 31}{18} = 0\ \ \ \ | \cdot 18\]

\[22x^{2} - 9x - 31 = 0\]

\[D = 81 + 4 \cdot 22 \cdot 31 = 2809\]

\[x = \frac{9 \pm \sqrt{2809}}{44} = \frac{9 \pm 53}{44}\]

\[x_{1} = \frac{31}{22} = 1\frac{9}{22},\ \ x_{2} = - 1\]

\[Ответ:x = 1\frac{9}{22};\ x = - 1.\]