Решебник по алгебре 9 класс Макарычев Задание 690

 

\[\boxed{\text{690}\text{\ (690)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = 2,\ \ d = 2,\]

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

\[2n = 200 \Longrightarrow n = 100;\]

\[S_{100} = \frac{a_{1} + a_{100}}{2} \cdot 100 = (2 + 200) \cdot 50 = 10100.\]

\[\textbf{б)}\ a_{1} = 1,\ \ d = 2,\ \ \]

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

\[2n - 1 = 149\]

\[2n = 150 \Longrightarrow n = 75,\]

\[S_{75} = \frac{a_{1} + a_{75}}{2} \cdot 75 = \frac{1 + 149}{2} \cdot 75 = 5625.\]

\[\textbf{в)}\ a_{1} = 102,\ \ d = 3,\]

\[a_{n} = a_{1} + d(n - 1) = 102 + 3n - 3 = 3n + 99 = 198\]

\[3n + 99 = 198\]

\[3n = 99\]

\[n = 33,\]

\[S_{33} = \frac{a_{1} + a_{33}}{2} \cdot 33 = \frac{102 + 198}{2} \cdot 33 = 150 \cdot 33 = 4950.\]

\[\boxed{\text{690.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \sqrt{19 - 6\sqrt{10}} = \sqrt{10} - 3\]

\[\sqrt{19 - 6\sqrt{10}} = \sqrt{10} - 3\]

\[\sqrt{10 - 2 \cdot 3\sqrt{10} + 9} = \sqrt{10} - 3\]

\[\sqrt{\left( \sqrt{10} - 3 \right)^{2}} = \sqrt{10} - 3\]

\[\sqrt{10} - 3 = \sqrt{10} - 3.\]

\[\textbf{б)}\ \sqrt{23 - 8\sqrt{7}} = 4 - \sqrt{7}\]

\[\sqrt{23 - 8\sqrt{7}} = 4 - \sqrt{7}\]

\[\sqrt{16 - 2 \cdot 4\sqrt{7} + 7} = 4 - \sqrt{7}\]

\[\sqrt{\left( 4 - \sqrt{7} \right)^{2}} = 4 - \sqrt{7}\]

\[4 - \sqrt{7} = 4 - \sqrt{7}.\]