Решебник по алгебре 9 класс Мерзляк Задание 857

 

\[\boxed{\mathbf{857\ (857).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[2x + 1,\ x + 5,\ x + 11\]

\[(x + 5)^{2} = (2x + 1)(x + 11)\ \]

\[x^{2} + 10x + 25 = 2x^{2} + 22x +\]

\[+ x + 11\]

\[x^{2} + 13x - 14 = 0\]

\[x_{1} + x_{2} = - 13;\ \ x_{1} = - 14\]

\[x_{1}x_{2} = - 14;\ \ x_{2} = 1\]

\[при\ x = 1:\ \ \]

\[2 \cdot 1 + 1 = 3,\ \ 1 + 5 = 6,\ \]

\[\ 1 + 11 = 12;\]

\[при\ x = - 14:\ \]

\[\ 2 \cdot ( - 14) + 1 = - 27,\]

\[\ \ - 14 + 5 = - 9,\ \ \]

\[- 14 + 11 = - 3.\]

\[Ответ:x = 1 \Longrightarrow b = 3;6;12\ \ или\ \]

\[\text{\ \ }x = - 14 \Longrightarrow b = - 27;\ - 9;\ - 3.\]

\[\boxed{\mathbf{857.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[Запишем\ числа:a = 3x + 1;.\]

\[Первое\ число\ a_{1} = 3 \cdot 0 + 1 = 1,\]

\[\text{\ \ }второе\ число\ a_{2} =\]

\[= 3 \cdot 2 + 1 = 7,\]

\[тогда\ \ \ d = 7 - 1 = 6.\]

\[S_{20} = \frac{a_{1} + a_{20}}{2} \cdot 20 =\]

\[= \frac{2a_{1} + 19d}{2} \cdot 20 =\]

\[= \frac{2 \cdot 1 + 19 \cdot 6}{2} \cdot 20 =\]

\[= (2 + 114) \cdot 10 = 1160.\]

\[Ответ:1160.\]