Задание
Найти значение выражения при \(\displaystyle x=5\):
\(\displaystyle (14+x)\cdot37+77\cdot(255:x)=\)
Решение
Подставим вместо \(\displaystyle x\) число \(\displaystyle 5\) в заданное выражение:
\(\displaystyle (14+x)\cdot 37+77\cdot (255:x)=(14+5)\cdot 37+77\cdot (255:5)=\,?\)
Расcтавим порядок действий:
| 1 | 3 | 5 | 4 | 2 | ||||||
| \(\displaystyle (14\) | \(\displaystyle +\) | \(\displaystyle 5)\) | \(\displaystyle \cdot\) | \(\displaystyle 37\) | \(\displaystyle +\) | \(\displaystyle 77\) | \(\displaystyle \cdot\) | \(\displaystyle (255\) | \(\displaystyle :\) | \(\displaystyle 5)\) |
1. Первое действие:
\(\displaystyle 14+5=19\).
2. Второе действие:
\(\displaystyle 255:5=51\).
3. Третье действие:
\(\displaystyle 19\cdot 37=703\).
| \(\displaystyle 1\) | \(\displaystyle 9\) | ||
| \(\displaystyle \times\) | |||
| \(\displaystyle 3\) | \(\displaystyle 7\) | ||
| \(\displaystyle 1\) | \(\displaystyle 3\) | \(\displaystyle 3\) | |
| \(\displaystyle +\) | |||
| \(\displaystyle 5\) | \(\displaystyle 7\) | ||
| \(\displaystyle 7\) | \(\displaystyle 0\) | \(\displaystyle 3\) | |
4. Четвёртое действие:
\(\displaystyle 77\cdot 51=3927\).
| \(\displaystyle 7\) | \(\displaystyle 7\) | |||
| \(\displaystyle \times\) | ||||
| \(\displaystyle 5\) | \(\displaystyle 1\) | |||
| \(\displaystyle 7\) | \(\displaystyle 7\) | |||
| \(\displaystyle +\) | ||||
| \(\displaystyle 3\) | \(\displaystyle 8\) | \(\displaystyle 5\) | ||
| \(\displaystyle 3\) | \(\displaystyle 9\) | \(\displaystyle 2\) | \(\displaystyle 7\) | |
5. Пятое действие:
\(\displaystyle 703+3927=4630\).
| \(\displaystyle 7\) | \(\displaystyle 0\) | \(\displaystyle 3\) | ||
| \(\displaystyle +\) | ||||
| \(\displaystyle 3\) | \(\displaystyle 9\) | \(\displaystyle 2\) | \(\displaystyle 7\) | |
| \(\displaystyle \bf4\) | \(\displaystyle \bf6\) | \(\displaystyle \bf3\) | \(\displaystyle \bf0\) | |
Таким образом,
\(\displaystyle (14+5)\cdot 37+77\cdot (255:5)=4630\).
Ответ: \(\displaystyle 4630\).