Решебник по алгебре 9 класс Макарычев Задание 786

 

\[\boxed{\text{786\ (786).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 5\sqrt{x} = 1\]

\[\sqrt{x} = \frac{1}{5}\]

\[x = \left( \frac{1}{5} \right)^{2}\]

\[x = \frac{1}{25}.\]

\[\textbf{б)}\ \sqrt{x - 4} = 15\]

\[x - 4 = 15^{2}\]

\[x - 4 = 225\]

\[x = 229.\]

\[\boxed{\text{786.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[a_{6} + a_{10} = 5,9\]

\[a_{12} - a_{4} = 2\]

\[a_{25} = ?\]

\[a_{6} + a_{10} = a_{1} + 5d + a_{1} + 9d =\]

\[= 2a_{1} + 14d,\]

\[2a_{1} + 14d = 5,9;\]

\[a_{12} - a_{4} =\]

\[= a_{1} + 11d - a_{1} - 3d = 8d,\]

\[8d = 2,\ \ d = \frac{1}{4} = 0,25;\]

\[2a_{1} + 14d = 5,9\]

\[2a_{1} + 14 \cdot 0,25 = 5,9\]

\[2a_{1} + 3,5 = 5,9\]

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

\[a_{1} = 1,2.\]

\[a_{25} = a_{1} + 24d =\]

\[= 1,2 + 24 \cdot 0,25 = 7,2.\]