Решебник по алгебре 9 класс Макарычев Задание 689

 

\[\boxed{\text{689}\text{\ (689)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = 10,\ \ a_{n} = 99,\ \ d = 1\]

\[a_{n} = a_{1} + d(n - 1)\]

\[99 = 10 + n - 1\]

\[99 = 9 + n\]

\[n = 90;\]

\[S_{90} = \frac{a_{1} + a_{90}}{2} \cdot 90 = (10 + 99) \cdot 45 = 4905.\]

\[\textbf{б)}\ a_{1} = 100,\ \ d = 1,\ \ a_{n} = a_{1} + d(n - 1),\]

\[999 = 100 + n - 1\]

\[999 = 99 + n\]

\[n = 900\]

\[S_{900} = \frac{a_{1} + a_{900}}{2} \cdot 900 = (100 + 999) \cdot 450 = 494\ 550.\]

\[\boxed{\text{689.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{\sqrt{\sqrt{18} - 3} \cdot \sqrt{\sqrt{18} + 3}}{\sqrt{6}} = \sqrt{1,5}\]

\[\frac{\sqrt{\sqrt{18} - 3} \cdot \sqrt{\sqrt{18} + 3}}{\sqrt{6}} = \sqrt{1,5}\]

\[\frac{\sqrt{\left( \left( \sqrt{18} \right)^{2} - 3^{2} \right)}}{\sqrt{6}} = \sqrt{1,5}\]

\[\frac{\sqrt{9}}{\sqrt{6}} = \sqrt{1,5}\]

\[\sqrt{\frac{3}{2}} = \sqrt{1,5}\]

\[\sqrt{1,5} = \sqrt{1,5}.\]

\[\textbf{б)}\ \frac{\sqrt{10}}{\sqrt{7 + \sqrt{24}} \cdot \sqrt{7 - \sqrt{24}}} = \sqrt{0,4}\]

\[\ \frac{\sqrt{10}}{\sqrt{7 + \sqrt{24}} \cdot \sqrt{7 - \sqrt{24}}} = \sqrt{0,4}\]

\[\frac{\sqrt{10}}{\sqrt{\left( 7^{2} - \left( \sqrt{24} \right)^{2} \right)}} = \sqrt{0,4}\]

\[\frac{\sqrt{10}}{\sqrt{49 - 24}} = \sqrt{0,4}\]

\[\sqrt{\frac{10}{25}} = \sqrt{0,4}\]

\[\sqrt{0,4} = \sqrt{0,4}.\]