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↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\def\Base{8} \def\BaseB{2} \def\BaseI{3} \def\Ia{\frac12} \def\Ib{\frac13} \def\Ic{\frac16} \def\tubun{\frac{3+2-1}{6}} \def\tubunK{\frac{4}{6}} \def\tubunY{\frac{2}{3}} \def\Kotae{2^2=4} % \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \large \begin{align*} & \colMM{red}{ かけ算}\colMM{green}{ わり算}\\ & \colBX{bisque}{$\Base$}^{\Ia}\,\colBX{mistyrose}{$\times$}\,\colBX{bisque}{$\Base$}^{\Ib} \,\colBX{palegreen}{$\div$} \colBX{bisque}{$\Base$}^{\Ic}\\ & \colMM{red}{ 足す}\colMM{green}{ 引く}\\ &= \colBX{bisque}{$\Base$}^{\Ia \,\colBX{mistyrose}{$\scriptsize +$}\, \Ib \,\colBX{palegreen}{$\scriptsize -$}\, \Ic}\\ \\ &= \Base^{\tubun}\\ \\ &= \Base^{\tubunK}\\ \\ &= \Base^{\tubunY}\\ & \colMM{red}{ \searrow ◯^{▲}}\\ &= \left(\BaseB^{\BaseI}\right)^{\tubunY}\\ \\ &= 2^{\BaseI \times \tubunY}\\ \\ &= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\def\Base{2} \def\BaseB{2} \def\BaseI{1} \def\Ia{\frac32} \def\Ib{\frac43} \def\Ic{\frac56} \def\tubun{\frac{9+8-5}{6}} \def\tubunK{\frac{12}{6}} \def\tubunY{2} \def\Kotae{4} % \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \large \begin{align*} & \colMM{red}{ かけ算}\colMM{green}{ わり算}\\ & \colBX{bisque}{$\Base$}^{\Ia}\,\colBX{mistyrose}{$\times$}\,\colBX{bisque}{$\Base$}^{\Ib} \,\colBX{palegreen}{$\div$} \colBX{bisque}{$\Base$}^{\Ic}\\ & \colMM{red}{ 足す}\colMM{green}{ 引く}\\ &= \colBX{bisque}{$\Base$}^{\Ia \,\colBX{mistyrose}{$\scriptsize +$}\, \Ib \,\colBX{palegreen}{$\scriptsize -$}\, \Ic}\\ \\ &= \Base^{\tubun}\\ \\ &= \Base^{\tubunK}\\ \\ &= \Base^{\tubunY} = \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\def\Base{3} \def\BaseB{3} \def\BaseI{1} \def\Ia{\frac12} \def\Ib{\frac56} \def\Ic{\frac13} \def\tubun{\frac{3-5+2}{6}} \def\tubunK{\frac{0}{6}} \def\tubunY{0} \def\Kotae{1} % \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \large \begin{align*} & \colMM{red}{ わり算}\colMM{green}{ かけ算}\\ & \colBX{bisque}{$\Base$}^{\Ia}\,\colBX{mistyrose}{$\div$}\,\colBX{bisque}{$\Base$}^{\Ib} \,\colBX{palegreen}{$\times$} \colBX{bisque}{$\Base$}^{\Ic}\\ & \colMM{red}{ 引く}\colMM{green}{ 足す}\\ &= \colBX{bisque}{$\Base$}^{\Ia \,\colBX{mistyrose}{$\scriptsize -$}\, \Ib \,\colBX{palegreen}{$\scriptsize +$}\, \Ic}\\ \\ &= \Base^{\tubun}\\ \\ &= \Base^{\tubunK}\\ \\ &= \Base^{\tubunY} = \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan