Решебник по алгебре 9 класс Мерзляк Задание 736

 

\[\boxed{\mathbf{736\ (736).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[a_{1} = 4;\ \ a_{6} = - 5\]

\[a_{6} = a_{1} + 5d \Longrightarrow \ \ 5d = a_{6} -\]

\[- a_{1}\ \Longrightarrow \ d = \frac{a_{6} - a_{1}}{5} =\]

\[= \frac{- 5 - 4}{5} = - 1,8\]

\[a_{2} = 4 - 1,8 = 2,2\]

\[a_{3} = 2,2 - 1,8 = 0,4\]

\[a_{4} = 0,4 - 1,8 = - 1,4\]

\[a_{5} = - 1,4 - 1,8 = - 3,2\]

\[Ответ:2,2;\ \ 0,4;\ - 1,4;\ - 3,2.\]

\[\boxed{\mathbf{736.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{\frac{3}{2}},\ 1,\ \sqrt{\frac{2}{3}},\ \ldots;\ \ \ \]

\[q = 1\ :\sqrt{\frac{3}{2}} = \sqrt{\frac{2}{3}}\]

\[S = \sqrt{\frac{3}{2}}\ :\left( 1 - \sqrt{\frac{2}{3}} \right) =\]

\[= \sqrt{\frac{3}{2}}\ :\left( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}} \right) =\]

\[= \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{2} \cdot \left( \sqrt{3} - \sqrt{2} \right)} =\]

\[= \frac{\sqrt{3} \cdot \sqrt{2} \cdot \sqrt{3} \cdot \left( \sqrt{3} + \sqrt{2} \right)}{\sqrt{2} \cdot \sqrt{2} \cdot \left( \sqrt{3} - \sqrt{2} \right)\left( \sqrt{3} + \sqrt{2} \right)} =\]

\[= \frac{\sqrt{6} \cdot \left( 3 + \sqrt{6} \right)}{2 \cdot (3 - 2)} = \frac{3\sqrt{6} + 6}{2};\]

\[2)\ \frac{\sqrt{2} + 1}{\sqrt{2} - 1},\frac{1}{2 - \sqrt{2}},\frac{1}{2},\ldots\ \ \ \]

\[\ q = \frac{1}{2 - \sqrt{2}}\ :\frac{\sqrt{2} + 1}{\sqrt{2} - 1} =\]

\[= \frac{\sqrt{2} - 1}{\left( 2 - \sqrt{2} \right)\left( \sqrt{2} + 1 \right)} =\]

\[= \frac{1}{\sqrt{2} \cdot \left( \sqrt{2} + 1 \right)} = \frac{1}{2 + \sqrt{2}};\]

\[S = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}\ :\]

\[:\left( 1 - \frac{1}{\sqrt{2} \cdot \left( \sqrt{2} + 1 \right)} \right) =\]

\[= \frac{\sqrt{2} + 1}{\sqrt{2} - 1}\ :\frac{2 + \sqrt{2} - 1}{2 + \sqrt{2}} =\]

\[= \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \cdot \frac{2 + \sqrt{2}}{\sqrt{2} + 1} =\]

\[= \frac{\left( 2 + \sqrt{2} \right)\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} + 1 \right)} =\]

\[= \frac{2\sqrt{2} + 2 + 2 + \sqrt{2}}{2 - 1} = 4 + 3\sqrt{2}.\]