Страница 67, ГДЗ по математике за 4 класс к самостоятельным и контрольным работам Лопатковой
Умножение и деление на двузначное и трёхзначное число. Самостоятельная работа № 24. Вариант 2
\[\begin{aligned} 500\,200 \overset{\textcolor{#1a7f37}{2}}{-} 124\,949 \mathrel{\overset{\textcolor{#1a7f37}{1}}{\vphantom{+}{:}}} 307 \overset{\textcolor{#1a7f37}{3}}{+} 124\,535 &= \boxed{\smash[b]{\underset{\hspace{4em}}{\textcolor{#e03131}{\mathbf{624\,328}}}}} \end{aligned}\] \[\begin{array}{l} \textcolor{#1a7f37}{\mathbf{1)}} \\ \begin{array}{l|l} \mathllap{\smash{\raisebox{-0.5em}{$-$}}\,}\overset{\mathclap{\textcolor{#1c7ed6}{\overgroup{\hphantom{\textstyle 1249}}}}}{1249}49 & \underline{\,307\,} \\ \underline{1228}\phantom{00} & \overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#e03131}{\mathbf{4}\kern-0.075em}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#e03131}{\mathbf{0}\kern-0.075em}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#e03131}{\mathbf{7}\kern-0.075em}} \\ \phantom{00}\mathllap{\smash{\raisebox{-0.5em}{$-$}}\,}2149 & \\ \phantom{00}\underline{2149} & \\ \phantom{00000}0 & \end{array} \end{array} \qquad \begin{array}{l} \textcolor{#1a7f37}{\mathbf{2)}} \\ \begin{array}{cr} & \vphantom{\overset{\bullet}{0}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{5}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{0}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{0}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{2}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{0}}0 \\ \smash{\raisebox{0.62em}{$-$}} & \vphantom{\overset{\bullet}{0}}407 \\ \hline & \textcolor{#e03131}{\mathbf{4}\kern-0.075em\mathbf{9}\kern-0.075em\mathbf{9}\kern-0.075em\mathbf{7}\kern-0.075em\mathbf{9}\kern-0.075em\mathbf{3}\kern-0.075em} \\ & \vphantom{\underline{0}} \\ & \vphantom{\underline{0}} \end{array} \end{array} \qquad \begin{array}{l} \textcolor{#1a7f37}{\mathbf{3)}} \\ \begin{array}{cr} & \vphantom{\overset{\bullet}{0}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{4}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{9}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{9}}\overset{\textcolor{#1c7ed6}{\bullet}}{\textcolor{#1c7ed6}{7}}93 \\ \smash{\raisebox{0.62em}{$+$}} & \vphantom{\overset{\bullet}{0}}124535 \\ \hline & \textcolor{#e03131}{\mathbf{6}\kern-0.075em\mathbf{2}\kern-0.075em\mathbf{4}\kern-0.075em\mathbf{3}\kern-0.075em\mathbf{2}\kern-0.075em\mathbf{8}\kern-0.075em} \\ & \vphantom{\underline{0}} \\ & \vphantom{\underline{0}} \end{array} \end{array}\]
\[\overset{\textcolor{#808080}{\scriptstyle 8\,185\text{ кг}}}{8\text{ т }185\text{ кг}} \overset{\kern0.25em\textcolor{#808080}{\scriptstyle 21\,148\text{ кг}}\kern0.25em}{+} \overset{\textcolor{#808080}{\scriptstyle 12\,963\text{ кг}}}{12\text{ т }963\text{ кг}} = \boxed{\smash[b]{\underset{\hspace{4em}}{\textcolor{#e03131}{\mathbf{21}}}}}\text{ т }\boxed{\smash[b]{\underset{\hspace{4em}}{\textcolor{#e03131}{\mathbf{148}}}}}\text{ кг}\] \[\overset{\textcolor{#808080}{\scriptstyle 5\,500\text{ см}}}{55\text{ м}} \overset{\kern0.25em\textcolor{#808080}{\scriptstyle 565\text{ см}}\kern0.25em}{-} \overset{\textcolor{#808080}{\scriptstyle 4\,935\text{ см}}}{49\text{ м }35\text{ см}} = \boxed{\smash[b]{\underset{\hspace{4em}}{\textcolor{#e03131}{\mathbf{5}}}}}\text{ м }\boxed{\smash[b]{\underset{\hspace{4em}}{\textcolor{#e03131}{\mathbf{65}}}}}\text{ см}\] \[\overset{\textcolor{#808080}{\scriptstyle 115\text{ ч}}}{4\text{ сут. }19\text{ ч}} \overset{\kern0.25em\textcolor{#808080}{\scriptstyle 444\text{ ч}}\kern0.25em}{+} \overset{\textcolor{#808080}{\scriptstyle 329\text{ ч}}}{13\text{ сут. }17\text{ ч}} = \boxed{\smash[b]{\underset{\hspace{2em}}{\textcolor{#e03131}{\mathbf{18}}}}}\text{ сут. }\boxed{\smash[b]{\underset{\hspace{2em}}{\textcolor{#e03131}{\mathbf{12}}}}}\text{ ч}\]
\(1)\ 37 - 28 = \textcolor{#e03131}{\mathbf{9}}\text{ (ч)}\)
\(2)\ 1\,080 : 9 = \textcolor{#e03131}{\mathbf{120}}\text{ (км/ч)}\)
\(3)\ 28 + 37 = \textcolor{#e03131}{\mathbf{65}}\text{ (ч)}\)
\(4)\ 120 \cdot 65 = \textcolor{#e03131}{\mathbf{7\,800}}\text{ (км)}\)
Ответ: \(\textcolor{#e03131}{\mathbf{7\,800}}\) км.
\(1)\ 50\,880 : 24 = \textcolor{#e03131}{\mathbf{2\,120}}\text{ (м)}\)
\(2)\ P = (2\,120 + 24) \cdot 2 = \textcolor{#e03131}{\mathbf{4\,288}}\text{ (м)}\)
Ответ: \(\textcolor{#e03131}{\mathbf{4\,288}}\) м.
