Ответы к учебнику по математике за 2 класс 1 часть Рудницкая. Вычитание двузначных чисел (общий случай). Номер №5

Вычисли.
(95 − 28) + 17
(50 − 18) − 23
100 − (60 + 34)
92 − (63 − 34)
(45 + 45) − 90
(52 + 48) − 96

Ответ

(95 − 28) + 17 = 84
(50 − 18) − 23 = 9
100 − (60 + 34) = 6
92 − (63 − 34) = 63
(45 + 45) − 90 = 0
(52 + 48) − 96 = 4
Вычисления:
(95 − 28) + 17= 67 + 17 = 84

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}95\\ 28\end{array}}\\ \begin{array}{r}67\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}+\end{array}\underline{\begin{array}{r}67\\ 17\end{array}}\\ \begin{array}{r}84\end{array}\end{array}$
(50 − 18) − 23 = 32 − 23 = 9

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}50\\ 18\end{array}}\\ \begin{array}{r}32\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}32\\ 23\end{array}}\\ \begin{array}{r}9\end{array}\end{array}$
100 − (60 + 34) = 100 − 94 = 6

$\begin{array}{r}\begin{array}{r}+\end{array}\underline{\begin{array}{r}60\\ 34\end{array}}\\ \begin{array}{r}94\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}100\\ 94\end{array}}\\ \begin{array}{r}6\end{array}\end{array}$
92 − (63 − 34) = 92 − 29 = 63

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}63\\ 34\end{array}}\\ \begin{array}{r}29\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}92\\ 29\end{array}}\\ \begin{array}{r}63\end{array}\end{array}$
(45 + 45) − 90 = 90 − 90 = 0

$\begin{array}{r}\begin{array}{r}+\end{array}\underline{\begin{array}{r}45\\ 45\end{array}}\\ \begin{array}{r}90\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}90\\ 90\end{array}}\\ \begin{array}{r}0\end{array}\end{array}$
(52 + 48) − 96 = 100 − 96 = 4

$\begin{array}{r}\begin{array}{r}+\end{array}\underline{\begin{array}{r}52\\ 48\end{array}}\\ \begin{array}{r}100\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}100\\ 96\end{array}}\\ \begin{array}{r}4\end{array}\end{array}$