Так как
\(\displaystyle \sqrt[3\, ] {-0,1029}=-\sqrt[3\, ] {0,1029}\) и \(\displaystyle \sqrt[3\, ] {-0{,}3}=-\sqrt[3\, ] {0{,}3} \,{\small ,}\)
ПравилоПри \(\displaystyle a < 0{\small }\) и \(\displaystyle k \in \N\)
\(\displaystyle \sqrt[2k+1]{a}=-\sqrt[2k+1]{-a}{\small .}\)
получаем
\(\displaystyle \frac{\sqrt[3\,]{-0,1029}}{\sqrt[3\,]{-0{,}3}}=\frac{-\sqrt[3\,]{0,1029}}{-\sqrt[3\,]{0{,}3}} =\frac{\sqrt[3\,]{0,1029}}{\sqrt[3\,]{0{,}3}}{\small.}\)
По свойству арифметического корня\(\displaystyle \frac{\sqrt[3\,] {0,1029}}{\sqrt[3\, ] {0{,}3}}=\sqrt[3\,] {\frac {0,1029}{0{,}3}}=\sqrt[3\, ] {0{,}343}{\small .}\)
\(\displaystyle \sqrt[n\,]{\frac{a}{b}}={\frac{\sqrt[n]{a}}{\sqrt[n]{b}}}{\small}\) при \(\displaystyle a \geqslant 0 {\small,}\)\(\displaystyle b > 0{\small,}\)\(\displaystyle n \in \N{\small/}\)
\(\displaystyle \sqrt[3\, ] {0{,}343}=0{,}7 {\small .}\)
ИнформацияТаблица степеней
| | С т е п е н и |
| \(\displaystyle \bf \color{blue}{2}\) | \(\displaystyle \bf \color{blue}{3}\) | \(\displaystyle \bf \color{blue}{4}\) | \(\displaystyle \bf \color{blue}{5}\) | \(\displaystyle \bf \color{blue}{6}\) | \(\displaystyle \bf \color{blue}{7}\) | \(\displaystyle \bf \color{blue}{8}\) |
Ч
и
с
л
а | \(\displaystyle \bf \color{blue}{2}\) | \(\displaystyle 4\) | \(\displaystyle 8\) | \(\displaystyle 16\) | \(\displaystyle 32\) | \(\displaystyle 64\) | \(\displaystyle 128\) | \(\displaystyle 256\) |
| \(\displaystyle \bf \color{blue}{3}\) | \(\displaystyle 9\) | \(\displaystyle 27\) | \(\displaystyle 81\) | \(\displaystyle 243\) | \(\displaystyle 729\) | \(\displaystyle 2187\) | \(\displaystyle 6561\) |
| \(\displaystyle \bf \color{blue}{4}\) | \(\displaystyle 16\) | \(\displaystyle 64\) | \(\displaystyle 256\) | \(\displaystyle 1024\) | \(\displaystyle 4096\) | \(\displaystyle 16384\) | \(\displaystyle \ldots\) |
| \(\displaystyle \bf \color{blue}{5}\) | \(\displaystyle 25\) | \(\displaystyle 125\) | \(\displaystyle 625\) | \(\displaystyle 3125\) | \(\displaystyle 15625\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) |
| \(\displaystyle \bf \color{blue}{6}\) | \(\displaystyle 36\) | \(\displaystyle 216\) | \(\displaystyle 1296\) | \(\displaystyle 7776\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) |
| \(\displaystyle \bf \color{blue}{7}\) | \(\displaystyle 49\) | \(\displaystyle 343\) | \(\displaystyle 2401\) | \(\displaystyle 16807\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) |
| \(\displaystyle \bf \color{blue}{8}\) | \(\displaystyle 64\) | \(\displaystyle 512\) | \(\displaystyle 4096\) | \(\displaystyle 32768\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) |
| \(\displaystyle \bf \color{blue}{9}\) | \(\displaystyle 81\) | \(\displaystyle 729\) | \(\displaystyle 6561\) | \(\displaystyle 59049\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) |
| \(\displaystyle \bf \color{blue}{10}\) | \(\displaystyle 100\) | \(\displaystyle 1000\) | \(\displaystyle 10000\) | \(\displaystyle 100000\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) | \(\displaystyle \ldots\) |
Значит,
\(\displaystyle \frac{\sqrt[3\,]{-0,1029}}{\sqrt[3\,]{-0{,}3}}=\frac{-\sqrt[3\,]{0,1029}}{-\sqrt[3\,]{0{,}3}}=\sqrt[3\,]{\frac {0,1029}{0{,}3}}=\sqrt[3\,] {0{,}343}=0{,}7 {\small .}\)
Ответ: \(\displaystyle 0{,}7 {\small .}\)