Ответы к учебнику по математике за 2 класс 1 часть Рудницкая. Взаимное расположение фигур на плоскости. Номер №17

Выполни вычитание.
(72 − 6) − 54
(32 − 8) − 12
90 − (36 − 29)
80 − (47 − 28)
(70 − 35) − (19 − 8)
(42 − 14) − (100 − 87)
(51 − 9) − (45 − 15)
(100 − 6) − (88 − 29)

Ответ

(72 − 6) − 54 = 12
(32 − 8) − 12 = 12
90 − (36 − 29) = 83
80 − (47 − 28) = 61
(70 − 35) − (19 − 8) = 24
(42 − 14) − (100 − 87) = 15
(51 − 9) − (45 − 15) = 12
(100 − 6) − (88 − 29) = 35
Вычисления:
(72 − 6) − 54 = 66 − 54 = 12

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}72\\ 6\end{array}}\\ \begin{array}{r}66\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}66\\ 54\end{array}}\\ \begin{array}{r}12\end{array}\end{array}$
(32 − 8) − 12 = 24 − 12 = 12

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}32\\ 8\end{array}}\\ \begin{array}{r}24\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}24\\ 12\end{array}}\\ \begin{array}{r}12\end{array}\end{array}$
90 − (36 − 29) = 90 − 7 = 83

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}36\\ 29\end{array}}\\ \begin{array}{r}7\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}90\\ 7\end{array}}\\ \begin{array}{r}83\end{array}\end{array}$
80 − (47 − 28) = 80 − 19 = 61

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}47\\ 28\end{array}}\\ \begin{array}{r}19\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}80\\ 19\end{array}}\\ \begin{array}{r}61\end{array}\end{array}$
(70 − 35) − (19 − 8) = 35 − 11 = 24

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}70\\ 35\end{array}}\\ \begin{array}{r}35\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}19\\ 8\end{array}}\\ \begin{array}{r}11\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}35\\ 11\end{array}}\\ \begin{array}{r}24\end{array}\end{array}$
(42 − 14) − (100 − 87) = 28 − 13 = 15

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}42\\ 14\end{array}}\\ \begin{array}{r}28\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}100\\ 87\end{array}}\\ \begin{array}{r}13\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}28\\ 13\end{array}}\\ \begin{array}{r}15\end{array}\end{array}$
(51 − 9) − (45 − 15) = 42 − 30 = 12

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}51\\ 9\end{array}}\\ \begin{array}{r}42\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}45\\ 15\end{array}}\\ \begin{array}{r}30\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}42\\ 30\end{array}}\\ \begin{array}{r}12\end{array}\end{array}$
(100 − 6) − (88 − 29) = 94 − 59 = 35

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}100\\ 6\end{array}}\\ \begin{array}{r}94\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}88\\ 29\end{array}}\\ \begin{array}{r}59\end{array}\end{array}$

$\begin{array}{r}\begin{array}{r}—\end{array}\underline{\begin{array}{r}94\\ 59\end{array}}\\ \begin{array}{r}35\end{array}\end{array}$