Определение Среднее геометрическое
Средним геометрическим \(\displaystyle n\) положительных чисел называется такое положительное число \(\displaystyle a\small,\) что число \(\displaystyle a^n\) равно произведению данных чисел.
Требуется найти среднее геометрическое двух чисел, значит \(\displaystyle n=2\small.\)
Произведением данных чисел \(\displaystyle 0{,}36\) и \(\displaystyle 0{,}81\) является число
\(\displaystyle 0{,}36 \cdot 0{,}81=0{,}2916\small.\)
Теперь требуется найти такое положительное число \(\displaystyle a\small,\) что число \(\displaystyle a^2\) равно \(\displaystyle 0{,}2916\small.\)
То есть надо найти положительное число \(\displaystyle a\small,\) квадрат которого равен \(\displaystyle 0{,}2916\small.\)
В таблице квадратов есть число \(\displaystyle 2916{\small .}\)
По таблице квадратов находим, что \(\displaystyle 2916=54^2{\small .}\)
| | Е д и н и ц ы |
| \(\displaystyle \bf \color{blue}{0}\) | \(\displaystyle \bf \color{blue}{1}\) | \(\displaystyle \bf \color{blue}{2}\) | \(\displaystyle \bf \color{blue}{3}\) | \(\displaystyle \bf \color{blue}{4}\) | \(\displaystyle \bf \color{blue}{5}\) | \(\displaystyle \bf \color{blue}{6}\) | \(\displaystyle \bf \color{blue}{7}\) | \(\displaystyle \bf \color{blue}{8}\) | \(\displaystyle \bf \color{blue}{9}\) |
Д
е
с
я
т
к
и | \(\displaystyle \bf \color{blue}{1}\) | \(\displaystyle 100\) | \(\displaystyle 121\) | \(\displaystyle 144\) | \(\displaystyle 169\) | \(\displaystyle 196\) | \(\displaystyle 225\) | \(\displaystyle 256\) | \(\displaystyle 289\) | \(\displaystyle 324\) | \(\displaystyle 361\) |
| \(\displaystyle \bf \color{blue}{2}\) | \(\displaystyle 400\) | \(\displaystyle 441\) | \(\displaystyle 484\) | \(\displaystyle 529\) | \(\displaystyle 576\) | \(\displaystyle 625\) | \(\displaystyle 676\) | \(\displaystyle 729\) | \(\displaystyle 784\) | \(\displaystyle 841\) |
| \(\displaystyle \bf \color{blue}{3}\) | \(\displaystyle 900\) | \(\displaystyle 961\) | \(\displaystyle 1024\) | \(\displaystyle 1089\) | \(\displaystyle 1156\) | \(\displaystyle 1225\) | \(\displaystyle 1296\) | \(\displaystyle 1369\) | \(\displaystyle 1444\) | \(\displaystyle 1521\) |
| \(\displaystyle \bf \color{blue}{4}\) | \(\displaystyle 1600\) | \(\displaystyle 1681\) | \(\displaystyle 1764\) | \(\displaystyle 1849\) | \(\displaystyle 1936\) | \(\displaystyle 2025\) | \(\displaystyle 2116\) | \(\displaystyle 2209\) | \(\displaystyle 2304\) | \(\displaystyle 2401\) |
| \(\displaystyle \bf \color{blue}{5}\) | \(\displaystyle 2500\) | \(\displaystyle 2601\) | \(\displaystyle 2704\) | \(\displaystyle 2809\) | \(\displaystyle 2916\) | \(\displaystyle 3025\) | \(\displaystyle 3136\) | \(\displaystyle 3249\) | \(\displaystyle 3364\) | \(\displaystyle 3481\) |
| \(\displaystyle \bf \color{blue}{6}\) | \(\displaystyle 3600\) | \(\displaystyle 3721\) | \(\displaystyle 3844\) | \(\displaystyle 3969\) | \(\displaystyle 4096\) | \(\displaystyle 4225\) | \(\displaystyle 4356\) | \(\displaystyle 4489\) | \(\displaystyle 4624\) | \(\displaystyle 4761\) |
| \(\displaystyle \bf \color{blue}{7}\) | \(\displaystyle 4900\) | \(\displaystyle 5041\) | \(\displaystyle 5184\) | \(\displaystyle 5329\) | \(\displaystyle 5476\) | \(\displaystyle 5625\) | \(\displaystyle 5776\) | \(\displaystyle 5929\) | \(\displaystyle 6084\) | \(\displaystyle 6241\) |
| \(\displaystyle \bf \color{blue}{8}\) | \(\displaystyle 6400\) | \(\displaystyle 6561\) | \(\displaystyle 6724\) | \(\displaystyle 6889\) | \(\displaystyle 7056\) | \(\displaystyle 7225\) | \(\displaystyle 7396\) | \(\displaystyle 7569\) | \(\displaystyle 7744\) | \(\displaystyle 7921\) |
| \(\displaystyle \bf \color{blue}{9}\) | \(\displaystyle 8100\) | \(\displaystyle 8281\) | \(\displaystyle 8464\) | \(\displaystyle 8649\) | \(\displaystyle 8836\) | \(\displaystyle 9025\) | \(\displaystyle 9216\) | \(\displaystyle 9409\) | \(\displaystyle 9604\) | \(\displaystyle 9801\) |
\(\displaystyle \color{Orange} {2916}=\color{Magenta}5 \blue 4^2{\small .}\)
Тогда
\(\displaystyle 0{,}2916=\frac{2916}{10000}=\frac{54^2}{100^2}= \left(\frac{54}{100}\right)^2=(0{,}54)^2{\small .}\)
Значит, среднее геометрическое чисел \(\displaystyle 0{,}36\) и \(\displaystyle 0{,}81\) – число \(\displaystyle 0{,}54{\small . } \)
Ответ: \(\displaystyle 0{,}54{\small . } \)