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МатематикаРешебник по алгебре 8 класс Мерзляк Задание 660
Задание 660
\[\boxed{\mathbf{660\ (660).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ x² - 3x + 2 = 0\]
\[D = 9 - 4 \cdot 2 = 9 - 8 = 1\]
\[x = \frac{3 \pm \sqrt{1}}{2} = \frac{3 \pm 1}{2}\]
\[x_{1} = 2\]
\[x_{2} = 1\]
\[Ответ:x = 1;x = 2.\]
\[2)\ x² + 12x - 13 = 0\]
\[D = 144 + 4 \cdot 13 = 196\]
\[x = \frac{- 12 \pm \sqrt{196}}{2} = \frac{- 12 \pm 14}{2}\]
\[x_{1} = 1\]
\[x_{2} = - 13\]
\[Ответ:\ x = - 13;x = 1.\]
\[3)\ x² - 7x + 10 = 0\]
\[D = 49 - 40 = 9\]
\[x = \frac{7 \pm \sqrt{9}}{2} = \frac{7 \pm 3}{2}\]
\[x_{1} = 5\]
\[x_{2} = 2\]
\[Ответ:x = 2;x = 5.\]
\[4)\ x² - x - 72 = 0\]
\[D = 1 + 4 \cdot 72 = 289\]
\[x = \frac{1 \pm \sqrt{289}}{2} = \frac{1 \pm 17}{2}\]
\[x_{1} = 9\]
\[x_{2} = - 8\]
\[Ответ:\ x = - 8;x = 9.\]
\[5)\ 2x² - 5x + 2 = 0\]
\[D = ( - 5)^{2} - 4 \cdot 2 \cdot 2 =\]
\[= 25 - 16 = 9\]
\[x_{1,2} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}\]
\[x_{1} = \frac{1}{2} = 0,5;\ \ \ x_{2} = 2\ \]
\[Ответ:x = 0,5;\ \ x = 2.\]
\[6)\ \ 2x² - 7x - 4 = 0\]
\[D = ( - 7)^{2} - 4 \cdot 2 \cdot ( - 4) =\]
\[= 49 + 32 = 81\]
\[x_{1,2} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4}\]
\[x_{1} = - \frac{1}{2} = - 0,5;\ \ x_{2} = 4\]
\[Ответ:\ x = - 0,5;x = 4.\]
\[7)\ 4x² - 3x - 1 = 0\]
\[D = 9 - 4 \cdot 4 \cdot ( - 1) = 25\]
\[x_{1,2} = \frac{3 \mp \sqrt{25}}{8} = \frac{3 \pm 5}{8}\]
\[x_{1} = - \frac{1}{4} = - 0,25;\ \ x_{2} = 1\]
\[Ответ:\ x = - 0,25;\ \ x = 1.\]
\[8) - 2x^{2} + x + 15 = 0\]
\[D = 1 - 4 \cdot ( - 2) \cdot 15 = 121\]
\[x_{1,2} = \frac{- 1 \pm \sqrt{121}}{- 4} = \frac{- 1 \pm 11}{- 4}\]
\[x_{1} = 3;\ \ x_{2} = - 2,5\]
\[Ответ:x = 3;\ x = - 2,5.\]
\[9)\ 6x² + 7x - 5 = 0\]
\[D = 49 - 4 \cdot 6 \cdot ( - 5) = 169\]
\[x_{1,2} = \frac{- 7 \pm \sqrt{169}}{12} = \frac{- 7 \pm 13}{12}\]
\[x_{1} = \frac{1}{2} = 0,5;\ \ \ x_{2} = - 1\frac{2}{3}\]
\[Ответ:x = 0,5;\ x = - 1\frac{2}{3}.\]
\[10)\ 18x² - 9x - 5 = 0\]
\[D = 81 - 4 \cdot 18 \cdot ( - 5) = 441\]
\[x_{1,2} = \frac{9 \pm \sqrt{441}}{36} = \frac{9 \pm 21}{36}\]
\[x_{1} = \frac{5}{6};\ \ \ x_{2} = - \frac{1}{3}\]
\[Ответ:x = \frac{5}{6};\ x = - \frac{1}{3}.\]
\[11)\ x² - 6x + 11 = 0\]
\[D = 36 - 4 \cdot 1 \cdot 11 = - 8 < 0\]
\[Ответ:нет\ корней.\]
\[12) - x^{2} - 8x + 12 = 0\]
\[D = 64 - 4 \cdot ( - 1) \cdot 12 = 112\]
\[x_{1,2} = \frac{8 \pm \sqrt{112}}{- 2} = \frac{8 \pm 4\sqrt{7}}{- 2} =\]
\[= - 4 \pm 2\sqrt{7}\]
\[Ответ:\ x = - 4 \pm 2\sqrt{7}.\]
