Решебник по алгебре 8 класс Мерзляк ФГОС Задание 698

 

\[\boxed{\mathbf{698\ (698).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\frac{\left( a^{- 3} \right)^{3}}{a^{- 2} \cdot a^{- 5}} = \frac{a^{- 9}}{a^{- 7}} = \frac{a^{7}}{a^{9}} = \frac{1}{a^{2}};\ \ \]

\[при\ a = \frac{1}{3}:\]

\[\frac{1}{a^{2}} = \frac{1}{\left( \frac{1}{3} \right)^{2}} = 1\ :\frac{1}{9} = 9.\]

\[\boxed{\mathbf{6}\mathbf{9}\mathbf{8}\mathbf{\text{.\ }}Еуроки - \ ДЗ\ без\ мороки}\]

\[1)\ 6x² - 2 = 5 - x\]

\[6x^{2} + x - 7 = 0\]

\[D = 1 + 4 \cdot 6 \cdot 7 = 169\]

\[x = \frac{- 1 \pm \sqrt{169}}{12} = \frac{- 1 \pm 13}{12}\]

\[x_{1} = 1\]

\[x_{2} = - \frac{7}{6} = - 1\frac{1}{6}\]

\[Ответ:x = 1;\ x = - 1\frac{1}{6}.\]

\[2)\ y - 6 = y^{2} - 9y + 3\]

\[- y^{2} + 10y - 9 = 0\]

\[y^{2} - 10y + 9 = 0\]

\[D = 100 - 36 = 64\]

\[y = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}\]

\[y_{1} = 9\]

\[y_{2} = 1\]

\[Ответ:y = 1;y = 9.\]

\[3)\ 4m² + 4m + 2 =\]

\[= 2m² + 10m + 8\]

\[2m^{2} - 6m - 6 = 0\]

\[D = 36 + 48 = 84\]

\[m = \frac{6 \pm 2\sqrt{21}}{2 \cdot 2} = \frac{3 \pm \sqrt{21}}{2}\]

\[Ответ:\ m = \frac{3 \pm \sqrt{21}}{2}.\]