Решебник по алгебре 8 класс Мерзляк ФГОС Задание 799

 

\[\boxed{\mathbf{799\ (799).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\frac{\sqrt{6} + 2}{\sqrt{6} - 2} - \frac{\sqrt{6} - 2}{\sqrt{6} + 2} =\]

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

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

\[= \frac{8\sqrt{6}}{2} = 4\sqrt{6}\]

\[Ответ:иррациональное.\]

\[\boxed{\mathbf{7}\mathbf{9}\mathbf{9}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1.\frac{25a^{2} - 36}{10a^{2} - 9a + 2} =\]

\[= \frac{(5a - 6)(5a + 6)}{10 \cdot (a - 0,4)(a - 0,5)}\]

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

\[a_{1} + a_{2} = 0,9,\ \ a_{1} = 0,4\]

\[a_{1} \cdot a_{2} = 0,2,\ \ a_{2} = 0,5\]

\[3.\frac{5a - 6}{2a - 1} - \frac{9a - 8}{2a - 1} =\]

\[= \frac{5a - 6 - 9a + 8}{2a - 1} = \frac{- 4a + 2}{2a - 1} =\]

\[= \frac{- 2 \cdot (2a - 1)}{(2a - 1)} = - 2\]

\[1.\frac{4}{a^{2} + 2a - 3} = \frac{4}{(a + 3)(a - 1)}\]

\[a² + 2a - 3 = 0\]

\[a_{1} + a_{2} = - 2,\ \ a_{1} = - 3\]

\[a_{1} \cdot a_{2} = - 3,\ \ a_{2} = 1\]

\[2.\frac{2a}{a + 3} + \frac{1}{a - 1} - \frac{4}{(a + 3)(a - 1)} =\]

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

\[= \frac{2a^{2} - a - 1}{(a + 3)(a - 1)} =\]

\[= \frac{2 \cdot (a + 0,5)(a - 1)}{(a + 3)(a - 1)}\]

\[2a² - a - 1 = 0\]

\[a_{1} + a_{2} = \frac{1}{2},\ \ a_{1} = - \frac{1}{2}\]

\[a_{1} \cdot a_{2} = - \frac{1}{2},\ \ a_{2} = 1\]

\[3.\frac{(2a + 1)(a - 1)(a + 3)}{(a + 3)(a - 1)(2a + 1)} = 1\]