Решебник по алгебре 8 класс Мерзляк ФГОС Задание 664

 

\[\boxed{\mathbf{664\ (664).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ (x - 4)^{2} = 4x - 11\]

\[x^{2} - 8x + 16 - 4x + 11 = 0\]

\[x^{2} - 12x + 27 = 0\]

\[D = 144 - 4 \cdot 27 = 144 - 108 =\]

\[= 36\]

\[x = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2}\]

\[x_{1} = 9,\ \ x_{2} = 3\]

\[Ответ:x = 3;x = 9.\]

\[2)\ (x + 5)^{2} + (x - 7)(x + 7) =\]

\[= 6x - 19\]

\[2x^{2} + 4x - 5 = 0\]

\[D = 16 + 40 = 56\]

\[x = \frac{- 4 \pm \sqrt{56}}{4} = \frac{- 4 \pm 2\sqrt{14}}{4} =\]

\[= \frac{- 2 \pm \sqrt{14}}{2}\]

\[Ответ:x = \frac{- 2 \pm \sqrt{14}}{2}\text{.\ }\]

\[3)\ (3x - 1)(x + 4) =\]

\[= (2x + 3)(x + 3) - 17\]

\[3x^{2} + 12x - x - 4 =\]

\[= 2x^{2} + 6x + 3x + 9 - 17\]

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

\[D = 4 - 16 < 0\]

\[Ответ:нет\ корней.\]

\[\boxed{\mathbf{6}\mathbf{6}\mathbf{4}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ (3x - 1)(x + 4) = - 4\]

\[3x^{2} + 12x - x - 4 + 4 = 0\]

\[3x² + 11x = 0\]

\[x(3x + 11) = 0\]

\[x = 0,\ \ \ \ 3x + 11 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - \frac{11}{3} = - 3\frac{2}{3}\]

\[Ответ:\ x = - 3\frac{2}{3};0.\]

\[2)\ (2x - 1)^{2} - 6 \cdot (6 - x) = 2x\]

\[4x^{2} - 4x + 1 - 36 + 6x - 2x =\]

\[= 0\]

\[4x^{2} - 6x + 6x - 35 = 0\]

\[4x^{2} = 35\]

\[x^{2} = \frac{35}{4}\]

\[x = \sqrt{\frac{35}{4}}\]

\[x = - \sqrt{\frac{35}{4}}\]

\[x = \frac{\sqrt{35}}{2}\]

\[x = - \frac{\sqrt{35}}{2}\]

\[Ответ:\ x = - \frac{\sqrt{35}}{2};\frac{\sqrt{35}}{2}.\]

\[x^{2} - x - 6 - x^{2} + 25 - x^{2} + x =\]

\[= 0\]

\[- x^{2} + 19 = 0\]

\[x^{2} = 19\]

\[x = \sqrt{19}\]

\[x = - \sqrt{19}\]

\[Ответ:\ x = - \sqrt{19};\sqrt{19}.\]