Найдите приближённое значение арифметического корня с одним знаком после запятой:
\(\displaystyle \sqrt{85}=\)\(\displaystyle ,\)\(\displaystyle \ldots\)
Шаг 1. Найдём целое число, с которого начинается десятичная запись \(\displaystyle \sqrt{85}{\small .}\)
Знаем, что \(\displaystyle 85\) лежит между двумя квадратами чисел:
\(\displaystyle 9^2<85<10^2{\small .}\)
Значит,
\(\displaystyle 9<\sqrt{85}<10{\small ,} \)
то есть
\(\displaystyle \sqrt{85}=9{,}\ldots\)
Шаг 2. Теперь найдём цифру в разряде десятых.
Чтобы получить более точное приближение, будем возводить в квадрат числа
\(\displaystyle 9{,}1{\small ;}\ \ 9{,}2{\small ;}\ \ldots{\small ,}\ 9{,}9\)
пока не получим число, большее \(\displaystyle 85{\small. }\)
Для вычислений воспользуемся таблицей квадратов.
Последний квадрат, меньший \(\displaystyle 85\)– это \(\displaystyle 84{,}64=\left(\green{9{,}2} \right)^2{\small .}\)
Первый квадрат, больший \(\displaystyle 85\)– это \(\displaystyle 86{,}49=\left(\red{9{,}3} \right)^2{\small .}\)
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 | \(\displaystyle \bf \color{blue}{1}\) | \(\displaystyle 100\) | \(\displaystyle 121\) | \(\displaystyle 144\) | \(\displaystyle 169\) | \(\displaystyle 196\) | \(\displaystyle 225\) | \(\displaystyle 256\) | \(\displaystyle 289\) | \(\displaystyle 324\) | \(\displaystyle 361\) |
| \(\displaystyle \bf \color{blue}{2}\) | \(\displaystyle 400\) | \(\displaystyle 441\) | \(\displaystyle 484\) | \(\displaystyle 529\) | \(\displaystyle 576\) | \(\displaystyle 625\) | \(\displaystyle 676\) | \(\displaystyle 729\) | \(\displaystyle 784\) | \(\displaystyle 841\) | |
| \(\displaystyle \bf \color{blue}{3}\) | \(\displaystyle 900\) | \(\displaystyle 961\) | \(\displaystyle 1024\) | \(\displaystyle 1089\) | \(\displaystyle 1156\) | \(\displaystyle 1225\) | \(\displaystyle 1296\) | \(\displaystyle 1369\) | \(\displaystyle 1444\) | \(\displaystyle 1521\) | |
| \(\displaystyle \bf \color{blue}{4}\) | \(\displaystyle 1600\) | \(\displaystyle 1681\) | \(\displaystyle 1764\) | \(\displaystyle 1849\) | \(\displaystyle 1936\) | \(\displaystyle 2025\) | \(\displaystyle 2116\) | \(\displaystyle 2209\) | \(\displaystyle 2304\) | \(\displaystyle 2401\) | |
| \(\displaystyle \bf \color{blue}{5}\) | \(\displaystyle 2500\) | \(\displaystyle 2601\) | \(\displaystyle 2704\) | \(\displaystyle 2809\) | \(\displaystyle 2916\) | \(\displaystyle 3025\) | \(\displaystyle 3136\) | \(\displaystyle 3249\) | \(\displaystyle 3364\) | \(\displaystyle 3481\) | |
| \(\displaystyle \bf \color{blue}{6}\) | \(\displaystyle 3600\) | \(\displaystyle 3721\) | \(\displaystyle 3844\) | \(\displaystyle 3969\) | \(\displaystyle 4096\) | \(\displaystyle 4225\) | \(\displaystyle 4356\) | \(\displaystyle 4489\) | \(\displaystyle 4624\) | \(\displaystyle 4761\) | |
| \(\displaystyle \bf \color{blue}{7}\) | \(\displaystyle 4900\) | \(\displaystyle 5041\) | \(\displaystyle 5184\) | \(\displaystyle 5329\) | \(\displaystyle 5476\) | \(\displaystyle 5625\) | \(\displaystyle 5776\) | \(\displaystyle 5929\) | \(\displaystyle 6084\) | \(\displaystyle 6241\) | |
| \(\displaystyle \bf \color{blue}{8}\) | \(\displaystyle 6400\) | \(\displaystyle 6561\) | \(\displaystyle 6724\) | \(\displaystyle 6889\) | \(\displaystyle 7056\) | \(\displaystyle 7225\) | \(\displaystyle 7396\) | \(\displaystyle 7569\) | \(\displaystyle 7744\) | \(\displaystyle 7921\) | |
| \(\displaystyle \bf \color{blue}{9}\) | \(\displaystyle 8100\) | \(\displaystyle 8281\) | \(\displaystyle 8464\) | \(\displaystyle 8649\) | \(\displaystyle 8836\) | \(\displaystyle 9025\) | \(\displaystyle 9216\) | \(\displaystyle 9409\) | \(\displaystyle 9604\) | \(\displaystyle 9801\) | |
Будем двигаться по первой строке слева направо:
\(\displaystyle \left(9{,}1 \right)^2= \left(\frac{91}{10}\right)^2=\frac{8281}{100}=82{,}81<85{\small, }\)
\(\displaystyle \left(9{,}2 \right)^2= \left(\frac{92}{10}\right)^2=\frac{8464}{100}=\green{84{,}64}<85{\small .}\)
\(\displaystyle \left(9{,}3\right)^2= \left(\frac{93}{10}\right)^2=\frac{8649}{100}=\red{86{,}49}>85{\small .}\)
Получаем:
\(\displaystyle \left(9{,}2 \right)^2< 85 < \left(9{,}3 \right)^2{\small .}\)
Тогда
\(\displaystyle 9{,}2<\sqrt{85}<9{,}3{\small.}\)
Значит,
\(\displaystyle \sqrt{85}=9{,}2\ldots\)
Ответ: \(\displaystyle \sqrt{85}=9{,}2\ldots\)