Решебник по алгебре 9 класс Макарычев Задание 884

\[\boxed{\text{884\ (884).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

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

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

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

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

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

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

\[\boxed{\text{884.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[1000 \cdot 21 = 21\ 000 = {144,9}^{2}\]

\[1001 \cdot 21 = 21\ 021 = {144,9}^{2}\]

\[1002 \cdot 21 = 21\ 042 = {145,1}^{2}\]

\[Подобным\ \ образом\ \]

\[проверяем\ остальные\ числа,\]

\[пока\ не\ дойдем\ до\ числа\ 1029:\]

\[1029 \cdot 21 = 21\ 609 = 147^{2}.\]

\[Ответ:1029.\]