指数法則〜有理数編〜

完成度２０％

授業で使うため，このページを作り始めたばかりです。したがって問題もほとんどありません。少しずつ問題を増やしていきます。ご期待ください。😞

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さっそく練習問題にチャレンジしましょう。

↓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

「ますどら」新着コンテンツ

累乗根はルートと同じ性質をもつヨ

完成度２０％

授業で使うため，このページを作り始めたばかりです。したがって問題もほとんどありません。少しずつ問題を増やしていきます。ご期待ください。😞

練習問題にチャレンジ♪

さっそく練習問題にチャレンジしましょう。

同じ累乗根の「かけ算」

↓この問題へのリンクはこちら（右クリックで保存）

\def\MondaiB{n}
\def\MondaiL{a}
\def\MondaiR{b}
\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}{\MondaiB 乗根と \MondaiB 乗根} & \colMM{red}{　　\Rightarrow １つの \MondaiB 乗根に}\\
\sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiL}\ \times \sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiR}
&= \sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiL \times \MondaiR}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{3}
\def\MondaiL{2}
\def\MondaiR{4}
\def\Keisan{8}
\def\KeisanB{2}
\def\KeisanI{3}
\def\Kotae{\\&=\KeisanB}
\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}{\MondaiB 乗根と \MondaiB 乗根} & \colMM{red}{　　\Rightarrow １つの \MondaiB 乗根に}\\
\sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiL}\ \times \sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiR}
&= \sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiL \times \MondaiR}\\
&= \sqrt[\MondaiB]{\Keisan}\\
&= \sqrt[\MondaiB]{\KeisanB^{\KeisanI}}
\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）

\def\MondaiB{n}
\def\MondaiL{a}
\def\MondaiR{b}
\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}{\MondaiB 乗根と \MondaiB 乗根} & \colMM{red}{　　\Rightarrow １つの \MondaiB 乗根に}\\
\sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiL}\ \times \sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiR}
&= \sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiL \times \MondaiR}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{3}
\def\MondaiL{3}
\def\MondaiR{9}
\def\Keisan{27}
\def\KeisanB{3}
\def\KeisanI{3}
\def\Kotae{\\&=\KeisanB}
\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}{\MondaiB 乗根と \MondaiB 乗根} & \colMM{red}{　　\Rightarrow １つの \MondaiB 乗根に}\\
\sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiL}\ \times \sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiR}
&= \sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiL \times \MondaiR}\\
&= \sqrt[\MondaiB]{\Keisan}\\
&= \sqrt[\MondaiB]{\KeisanB^{\KeisanI}}
\Kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

同じ累乗根の「わり算」

↓この問題へのリンクはこちら（右クリックで保存）

\def\MondaiB{n}
\def\MondaiL{a}
\def\MondaiR{b}
\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}{\MondaiB 乗根と \MondaiB 乗根} & \colMM{red}{　　\Rightarrow １つの \MondaiB 乗根に}\\
\dfrac{\sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiL}}{\sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiR}}
&= \sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\dfrac{\MondaiL}{\MondaiR}}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{3}
\def\MondaiU{12}
\def\MondaiD{3}
\def\Keisan{4}
\def\KeisanB{3}
\def\KeisanI{3}
\def\Kotae{\\&=\KeisanB}
\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}{\MondaiB 乗根と \MondaiB 乗根} & \colMM{red}{　　\Rightarrow １つの \MondaiB 乗根に}\\
\dfrac{\sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiU}}{\sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiD}}
&= \sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\dfrac{\MondaiU}{\MondaiD}}\\
&= \sqrt[\MondaiB]{\Keisan}\\
\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）

\def\MondaiB{n}
\def\MondaiL{a}
\def\MondaiR{b}
\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}{\MondaiB 乗根と \MondaiB 乗根} & \colMM{red}{　　\Rightarrow １つの \MondaiB 乗根に}\\
\dfrac{\sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiL}}{\sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\MondaiR}}
&= \sqrt[\colBX{mistyrose}{$\small\MondaiB$}]{\dfrac{\MondaiL}{\MondaiR}}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{4}
\def\MondaiU{32}
\def\MondaiD{2}
\def\Keisan{16}
\def\KeisanB{2}
\def\KeisanI{4}
\def\Kotae{\\&=\KeisanB}
\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}{\MondaiB 乗根と \MondaiB 乗根} & \colMM{red}{　　\Rightarrow １つの \MondaiB 乗根に}\\
\dfrac{\sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiU}}{\sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\MondaiD}}
&= \sqrt[\colBX{mistyrose}{$\scriptsize\MondaiB$}]{\dfrac{\MondaiU}{\MondaiD}}\\
\\
&= \sqrt[\MondaiB]{\Keisan}\\
\\
&= \sqrt[\MondaiB]{\KeisanB^{\KeisanI}}\\
\\
&= \KeisanB
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

累乗根のキホン

完成度２０％

授業で使うため，このページを作り始めたばかりです。したがって問題もほとんどありません。少しずつ問題を増やしていきます。ご期待ください。😞

練習問題にチャレンジ♪

さっそく練習問題にチャレンジしましょう。

↓この問題へのリンクはこちら（右クリックで保存）

\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}}}
\begin{align*}
\colMM{red}{\Darr n乗根} &\\
\sqrt[\colBX{mistyrose}{$\scriptsize n$}]{a^{\colBX{mistyrose}{$\scriptsize n$}}} &= a\colMM{red}{　　\Leftarrow とれる！}\\
\colMM{red}{n乗\Uarr}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan
\def\MondaiR{3}
\def\Mondai{8}
\def\MondaiB{2}
\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}}}
\begin{align*}
\colMM{orange}{◯^{▲}になおす！}& \colMM{red}{　　\Darr \MondaiR 乗根}\\
\sqrt[\MondaiR]{\Mondai} &= \sqrt[\colBX{mistyrose}{$\tiny\MondaiR$}]{\MondaiB^{\colBX{mistyrose}{$\tiny\MondaiR$}}}\\
& \colMM{red}{　　　\MondaiR乗 \Uarr}\\
&= \MondaiB
\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）

\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}}}
\begin{align*}
\colMM{red}{\Darr n乗根} &\\
\sqrt[\colBX{mistyrose}{$\scriptsize n$}]{a^{\colBX{mistyrose}{$\scriptsize n$}}} &= a\colMM{red}{　　\Leftarrow とれる！}\\
\colMM{red}{n乗\Uarr}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan
\def\MondaiR{4}
\def\Mondai{81}
\def\MondaiB{2}
\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}}}
\begin{align*}
\colMM{orange}{◯^{▲}になおす！}& \colMM{red}{　　\Darr \MondaiR 乗根}\\
\sqrt[\MondaiR]{\Mondai} &= \sqrt[\colBX{mistyrose}{$\tiny\MondaiR$}]{\MondaiB^{\colBX{mistyrose}{$\tiny\MondaiR$}}}\\
& \colMM{red}{　　　\MondaiR乗 \Uarr}\\
&= \MondaiB
\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）

\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}}}
\begin{align*}
\colMM{red}{\Darr n乗根} &\\
\sqrt[\colBX{mistyrose}{$\scriptsize n$}]{a^{\colBX{mistyrose}{$\scriptsize n$}}} &= a\colMM{red}{　　\Leftarrow とれる！}\\
\colMM{red}{n乗\Uarr}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan
\def\MondaiR{3}
\def\Mondai{1}
\def\MondaiB{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}}}
\begin{align*}
\colMM{orange}{◯^{▲}になおす！}& \colMM{red}{　　\Darr \MondaiR 乗根}\\
\sqrt[\MondaiR]{\Mondai} &= \sqrt[\colBX{mistyrose}{$\tiny\MondaiR$}]{\MondaiB^{\colBX{mistyrose}{$\tiny\MondaiR$}}}\\
& \colMM{red}{　　　\MondaiR乗 \Uarr}\\
&= \MondaiB
\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）

\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}}}
\begin{align*}
\colMM{red}{\Darr n乗根} &\\
\sqrt[\colBX{mistyrose}{$\scriptsize n$}]{a^{\colBX{mistyrose}{$\scriptsize n$}}} &= a\colMM{red}{　　\Leftarrow とれる！}\\
\colMM{red}{n乗\Uarr}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan
\def\MondaiR{3}
\def\Mondai{27}
\def\MondaiB{3}
\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}}}
\begin{align*}
\colMM{orange}{◯^{▲}になおす！}& \colMM{red}{　　\Darr \MondaiR 乗根}\\
\sqrt[\MondaiR]{\Mondai} &= \sqrt[\colBX{mistyrose}{$\tiny\MondaiR$}]{\MondaiB^{\colBX{mistyrose}{$\tiny\MondaiR$}}}\\
& \colMM{red}{　　　\MondaiR乗 \Uarr}\\
&= \MondaiB
\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）

\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}}}
\begin{align*}
\colMM{red}{\Darr n乗根} &\\
\sqrt[\colBX{mistyrose}{$\scriptsize n$}]{a^{\colBX{mistyrose}{$\scriptsize n$}}} &= a\colMM{red}{　　\Leftarrow とれる！}\\
\colMM{red}{n乗\Uarr}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan
\def\MondaiR{4}
\def\Mondai{\dfrac{1}{16}}
\def\MondaiB{\dfrac{1}{2}}
\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}}}
\begin{align*}
\colMM{orange}{◯^{▲}になおす！}& \colMM{red}{　　\Darr \MondaiR 乗根}\\
\sqrt[\MondaiR]{\Mondai} &= \sqrt[\colBX{mistyrose}{$\tiny\MondaiR$}]{\left(\MondaiB\right)^{\colBX{mistyrose}{$\tiny\MondaiR$}}}\\
& \colMM{red}{　　　\MondaiR乗 \Uarr}\\
&= \MondaiB
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

指数法則〜整数編〜

完成度２０％

授業で使うため，このページを作り始めたばかりです。したがって問題もほとんどありません。少しずつ問題を増やしていきます。ご期待ください。😞

練習問題にチャレンジ♪

さっそく練習問題にチャレンジしましょう。

かけ算は足し算に

↓この問題へのリンクはこちら（右クリックで保存）

\def\MondaiB{a}
\def\MondaiIl{m}
\def\MondaiIr{n}
\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}}}
\begin{align*}
\colMM{red}{かけ算は　} & \colMM{red}{　　\ 足し算に}\\
\MondaiB^{\MondaiIl} \colBX{mistyrose}{$\times$} \MondaiB^{\MondaiIr} &= \MondaiB\,^{\MondaiIl \colBX{mistyrose}{$\scriptsize +$} \MondaiIr}\\
\colMM{red}{底が同じ　}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{3}
\def\MondaiIl{5}
\def\MondaiIr{-2}
\def\KeisanI{3}
\def\Kotae{27}
\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}}}
\begin{align*}
\colMM{red}{かけ算は　} & \colMM{red}{　　\ 足し算に}\\
\MondaiB^{\MondaiIl} \colBX{mistyrose}{$\times$} \MondaiB^{\MondaiIr} &= \MondaiB\,^{\MondaiIl \colBX{mistyrose}{$\scriptsize +$} (\MondaiIr)}\\
\colMM{red}{底が同じ　}\\
&= 3^{\KeisanI}\\
\\
&= \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）

\def\MondaiB{a}
\def\MondaiIl{m}
\def\MondaiIr{n}
\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}}}
\begin{align*}
\colMM{red}{かけ算は　} & \colMM{red}{　　\ 足し算に}\\
\MondaiB^{\MondaiIl} \colBX{mistyrose}{$\times$} \MondaiB^{\MondaiIr} &= \MondaiB\,^{\MondaiIl \colBX{mistyrose}{$\scriptsize +$} \MondaiIr}\\
\colMM{red}{底が同じ　}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{3}
\def\MondaiIl{4}
\def\MondaiIr{-2}
\def\KeisanI{2}
\def\Kotae{9}
\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}}}
\begin{align*}
\colMM{red}{かけ算は　} & \colMM{red}{　　\ 足し算に}\\
\MondaiB^{\MondaiIl} \colBX{mistyrose}{$\times$} \MondaiB^{\MondaiIr} &= \MondaiB\,^{\MondaiIl \colBX{mistyrose}{$\scriptsize +$} (\MondaiIr)}\\
\colMM{red}{底が同じ　}\\
&= 3^{\KeisanI}\\
\\
&= \Kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

わり算は引き算に

↓この問題へのリンクはこちら（右クリックで保存）

\def\MondaiB{a}
\def\MondaiIl{m}
\def\MondaiIr{n}
\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}}}
\begin{align*}
\colMM{red}{わり算は　} & \colMM{red}{　　\ ひき算に}\\
\dfrac{\MondaiB^{\MondaiIl}}{\MondaiB^{\MondaiIr}} = \MondaiB^{\MondaiIl} \colBX{mistyrose}{$\div$} \MondaiB^{\MondaiIr} &= \MondaiB\,^{\MondaiIl \colBX{mistyrose}{$\scriptsize -$} \MondaiIr}\\
\colMM{red}{底が同じ　}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{3}
\def\MondaiIl{5}
\def\MondaiIr{-2}
\def\KeisanI{7}
\def\Kotae{\ \color{lightgray}{= 2187}}
\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}}}
\begin{align*}
\colMM{red}{わり算は　} & \colMM{red}{　　\ ひき算に}\\
\dfrac{\MondaiB^{\MondaiIl}}{\MondaiB^{\MondaiIr}} = \MondaiB^{\MondaiIl} \colBX{mistyrose}{$\div$} \MondaiB^{\MondaiIr} &= \MondaiB\,^{\MondaiIl \colBX{mistyrose}{$\scriptsize -$} (\MondaiIr)}\\
\colMM{red}{底が同じ　}\\
&= 3^{\KeisanI}\\
\\
& \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）

\def\MondaiB{a}
\def\MondaiIl{m}
\def\MondaiIr{n}
\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}}}
\begin{align*}
\colMM{red}{わり算は　} & \colMM{red}{　　\ ひき算に}\\
\dfrac{\MondaiB^{\MondaiIl}}{\MondaiB^{\MondaiIr}} = \MondaiB^{\MondaiIl} \colBX{mistyrose}{$\div$} \MondaiB^{\MondaiIr} &= \MondaiB\,^{\MondaiIl \colBX{mistyrose}{$\scriptsize -$} \MondaiIr}\\
\colMM{red}{底が同じ　}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{10}
\def\MondaiIl{-3}
\def\MondaiIr{2}
\def\KeisanI{-5}
\def\Kotae{\ =\dfrac{1}{10^{5}}\\\\ &= \dfrac{1}{100000} }
\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}}}
\begin{align*}
\colMM{red}{わり算は　} & \colMM{red}{　　\ ひき算に}\\
\dfrac{\MondaiB^{\MondaiIl}}{\MondaiB^{\MondaiIr}} = \MondaiB^{\MondaiIl} \colBX{mistyrose}{$\div$} \MondaiB^{\MondaiIr} &= \MondaiB\,^{\MondaiIl \colBX{mistyrose}{$\scriptsize -$} \MondaiIr}\\
\colMM{red}{底が同じ　}\\
&= \MondaiB^{\KeisanI}\\
\\
& \Kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

指数が並んだら「かける」

↓この問題へのリンクはこちら（右クリックで保存）

【解答】

\def\MondaiB{3}
\def\MondaiIa{5}
\def\MondaiIb{-2}
\def\KakkoLl{}\def\KakkoLr{}
\def\KakkoRl{(}\def\KakkoRr{)}
\def\KeisanI{-10}
\def\Kotae{\\&\colMM{lightgray}{　　\Darr 計算してみた}\\&\color{lightgray}=\dfrac{1}{3^{10}} = \dfrac{1}{59049}}
\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}}}
\begin{align*}
\colMM{red}{指数が並んだ} & \colMM{red}{　　　かける！}\\
\left(\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiIa$}}\right)^{\colBX{mistyrose}{$\scriptsize\MondaiIb$}} &= \MondaiB^{\colBX{mistyrose}{$\scriptsize\KakkoLl\MondaiIa\KakkoLr \times \KakkoRl\MondaiIb\KakkoRr$}}\\
\\
&= \MondaiB^{\KeisanI}
\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\MondaiB{3}
\def\MondaiIa{-2}
\def\MondaiIb{4}
\def\KakkoLl{}\def\KakkoLr{}
\def\KakkoRl{}\def\KakkoRr{}
\def\KeisanI{-8}
\def\Kotae{\\&\colMM{lightgray}{　　\Darr 計算してみた}\\&\color{lightgray}=\dfrac{1}{3^{8}} = \dfrac{1}{6561}}
\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}}}
\begin{align*}
\colMM{red}{指数が並んだ} & \colMM{red}{　　　かける！}\\
\left(\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiIa$}}\right)^{\colBX{mistyrose}{$\scriptsize\MondaiIb$}} &= \MondaiB^{\colBX{mistyrose}{$\scriptsize\KakkoLl\MondaiIa\KakkoLr \times \KakkoRl\MondaiIb\KakkoRr$}}\\
\\
&= \MondaiB^{\KeisanI}
\Kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

指数を拡張しよう〜整数編〜

完成度２０％

授業で使うため，このページを作り始めたばかりです。したがって問題もほとんどありません。少しずつ問題を増やしていきます。ご期待ください。😞

練習問題にチャレンジ♪

さっそく練習問題にチャレンジしましょう。

０乗は１・マイナス乗は逆数に

↓この問題へのリンクはこちら（右クリックで保存）

\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}{０乗は　　} & \colMM{red}{　　　\colFR{blue}{\color{red}\bf底に関係なく}常に１}\\
\colFR{blue}{$a$}^{\colBX{mistyrose}{$\small 0$}} &= \colBX{mistyrose}{$1$}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{2}
\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}}}
\begin{align*}
\colMM{red}{０乗は　　} & \colMM{red}{　　\colFR{blue}{\color{red}\bf底に関係なく}常に１}\\
\colFR{blue}{$\MondaiB$}^{\colBX{mistyrose}{$\small 0$}} &= \colBX{mistyrose}{$1$}\\
\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）

\def\MondaiB{a}
\def\MondaiI{-n}
\def\MondaiIm{n}
\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}}}
\begin{align*}
\colMM{red}{マイナス乗は　} & \colMM{red}{　　逆数に}\\
\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiI$}} &= \dfrac{1}{\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiIm$}}}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{10}
\def\MondaiI{-4}
\def\MondaiIm{4}
\def\Kotae{\dfrac{1}{10000}}
\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}}}
\begin{align*}
\colMM{red}{マイナス乗は　} & \colMM{red}{　　逆数に}\\
\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiI$}} &= \dfrac{1}{\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiIm$}}}\\
& \colMM{gray}{　　　\Darr \MondaiB^{\MondaiIm}\ は計算⁉}\\
&= \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）

\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}{０乗は　　} & \colMM{red}{　　　\colFR{blue}{\color{red}\bf底に関係なく}常に１}\\
\colFR{blue}{$a$}^{\colBX{mistyrose}{$\small 0$}} &= \colBX{mistyrose}{$1$}\\
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{5}
\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}}}
\begin{align*}
\colMM{red}{０乗は　　} & \colMM{red}{　　\colFR{blue}{\color{red}\bf底に関係なく}常に１}\\
\colFR{blue}{$\MondaiB$}^{\colBX{mistyrose}{$\small 0$}} &= \colBX{mistyrose}{$1$}\\
\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）

\def\MondaiB{a}
\def\MondaiI{-n}
\def\MondaiIm{n}
\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}}}
\begin{align*}
\colMM{red}{マイナス乗は　} & \colMM{red}{　　逆数に}\\
\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiI$}} &= \dfrac{1}{\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiIm$}}}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\MondaiB{4}
\def\MondaiI{-2}
\def\MondaiIm{2}
\def\Kotae{\dfrac{1}{16}}
\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}}}
\begin{align*}
\colMM{red}{マイナス乗は　} & \colMM{red}{　　逆数に}\\
\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiI$}} &= \dfrac{1}{\MondaiB^{\colBX{mistyrose}{$\scriptsize\MondaiIm$}}}\\
& \colMM{gray}{　　　\Darr \MondaiB^{\MondaiIm}\ は計算⁉}\\
&= \Kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan