Решебник по алгебре 8 класс Мерзляк ФГОС Задание 404

 

\[\boxed{\text{404\ (404).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ (2x - 3)^{2} = 25\]

\[(2x - 3)^{2} - 5^{2} = 0\]

\[(2x - 3 - 5)(2x - 3 + 5) = 0\]

\[(2x - 8)(2x + 2) = 0\]

\[2x = 8;\ \ \ \ \ \ 2x = - 2\]

\[x = 4\ \ \ \ \ \ \ \ \ \ x = - 1\]

\[Ответ:x = 4;x = \ - 1.\]

\[2)\ (x - 3)^{2} = 7\]

\[(x - 3)^{2} - \left( \sqrt{7} \right)^{2} = 0\]

\[\left( x - 3 - \sqrt{7} \right)\left( x - 3 + \sqrt{7} \right) = 0\]

\[x = 3 + \sqrt{7};\ \ \ \ \ x = 3 - \sqrt{7}\]

\[Ответ:x = \ \sqrt{7} + 3;\ x =\]

\[= - \sqrt{7} + 3.\]

\[3)\ (2x - 3)^{2} = 7\]

\[(2x - 3)^{2} - \left( \sqrt{7} \right)^{2} = 0\]

\[\left( 2x - 3 - \sqrt{7} \right)\left( 2x - 3 + \sqrt{7} \right) = 0\]

\[2x = 3 + \sqrt{7};\ \ \ \ \ \ \ \ 2x = 3 - \sqrt{7}\]

\[x = \frac{3 + \sqrt{7}}{2};\ \ \ \ \ \ \ \ \ x = \frac{3 - \sqrt{7}}{2}\]

\[Ответ:x = \frac{3 \pm \sqrt{7}}{2}\text{.\ }\]

\[\boxed{\text{404.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ 4^{2} = 16\]

\[2)\ 0^{2} = 0\]

\[3)\ {0,8}^{2} = 0,64\]

\[4)\ \left( 2\frac{1}{4} \right)^{2} = \left( \frac{9}{4} \right)^{2} = \frac{81}{16} = 5\frac{1}{16}\]

\[5)\ {1,6}^{2} = 2,56\]

\[6) - 9 < 0 \Longrightarrow не\ может\ быть\ \]

\[арифметическим\ \]

\[квадратным\ корнем.\ \]