Решебник по алгебре 9 класс Мерзляк Задание 765

 

\[\boxed{\mathbf{765\ (765).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[a_{1} = - 6;\ \ d = 4\]

\[S_{n} = \frac{2a_{1} + d(n - 1)}{2} \cdot n\]

\[S_{12} = \frac{2a_{1} + 11d}{2} \cdot 12 =\]

\[= \frac{- 12 + 44}{2} \cdot 12 = 32 \cdot 6 = 192.\]

\[\boxed{\mathbf{765.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{1}{x^{2} + 2x} - \frac{2}{x^{2} - 4} = \frac{x + 4}{5x(2 - x)}\]

\[\frac{1^{\backslash 5(x - 2)}}{x(x + 2)} - \frac{2^{\backslash 5x}}{(x - 2)(x + 2)} + \frac{x + 4^{\backslash x + 2}}{5x(x - 2)} = 0\]

\[ОДЗ:\ \ x \neq 0;\ \ x \neq \pm 2.\]

\[5x - 10 - 10x + x^{2} + 4x + 2x + 8 = 0\]

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

\[D = 1 + 8 = 9\]

\[x_{1} = \frac{- 1 + 3}{2} = 1;\]

\[x_{2} = \frac{- 1 - 3}{2} = - 2\ (не\ подходит).\]

\[Ответ:\ \ x = 1.\]

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

\[\frac{2^{\backslash x^{2} + x + 1}}{(x - 1)^{2}} - \frac{1^{\backslash x - 1}}{(x - 1)\left( x^{2} + x + 1 \right)} - \frac{3^{\backslash\text{(}x - 1)²}}{x^{2} + x + 1} = 0\]

\[ОДЗ:\ \ x \neq 1.\]

\[2x^{2} + 2x + 2 - x + 1 - 3\left( x^{2} - 2x + 1 \right) = 0\]

\[2x^{2} + x + 3 - 3x^{2} + 6x - 3 = 0\]

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

\[- x(x - 7) = 0\]

\[x = 0;\ \ x = 7.\]

\[Ответ:\ \ x = 0;x = 7.\]