Решебник по алгебре 8 класс Мерзляк Задание 671

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\[1)\ x² + 3x\sqrt{2} + 4 = 0\]

\[x^{2} + 3\sqrt{2} \cdot x + 4 = 0\]

\[D = \left( 3\sqrt{2} \right)^{2} - 16 = 18 - 16 = 2\]

\[x = \frac{- 3\sqrt{2} \pm \sqrt{2}}{2}\]

\[x_{1} = \frac{- 4\sqrt{2}}{2} = - 2\sqrt{2}\]

\[x_{2} = \frac{- 2\sqrt{2}}{2} = - \sqrt{2}\]

\[Ответ:\ x = - 2\sqrt{2};\ x = - \sqrt{2}.\]

\[2)\ x² - x\left( \sqrt{3} + 2 \right) + 2\sqrt{3} = 0\]

\[D = \left( \sqrt{3} + 2 \right)^{2} - 8\sqrt{3} =\]

\[= 3 + 4\sqrt{3} + 4 - 8\sqrt{3} =\]

\[= 7 - \sqrt{3} = \left( 2 - \sqrt{3} \right)^{2}\]

\[x = \frac{\left( \sqrt{3} + 2 \right) \pm \sqrt{\left( 2 - \sqrt{3} \right)^{2}}}{2} =\]

\[= \frac{\left( \sqrt{3} + 2 \right) \pm \left( 2 - \sqrt{3} \right)}{2}\]

\[x_{1} = 2,\ \ x_{2} = \sqrt{3}\]

\[Ответ:x = 2;x = \sqrt{3}.\]

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

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

\[8x^{2} - 11x + 3 = 0\]

\[D = 121 - 4 \cdot 8 \cdot 3 = 25\]

\[x = \frac{11 \pm \sqrt{25}}{16} = \frac{11 \pm 5}{16}\]

\[x_{1} = 1,\ \ x_{2} = \frac{3}{8}\]

\[Ответ:x = \frac{3}{8};x = 1.\]