平方根の積と商
a>0,b>0 のとき
\sqrt{a}\sqrt{b} = \sqrt{ab}
\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{\fbox{Point} } & \ \colMM{red}{\sqrt{a}\sqrt{b}\ \ \ \ \ が\ \ \ ab\ の正の平方根}\\ & \color{red}\sqrt{a}\sqrt{b} = +\sqrt{ab}\\ & \colMM{red}{ であることを示す!}\\ \\ \colMM{black}{\fbox{証明} } & \colMM{orange}{2乗したら\ ab\ ➡\ ab\ の平方根!}\\ \left(\sqrt{a}\sqrt{b}\right)^{\colBX{bisque}{$\scriptsize 2$}} &= (\sqrt{a})^2(\sqrt{b})^2 = \colFR{red}{$ab$}\\ \\ そして, & \color{lightgray}a>0,\ b>0\ より\\ \\ & \sqrt{a}>0,\ \sqrt{b}>0\ より\\ & \colMM{green}{ \swarrow\sqrt{a}\sqrt{b}\ は正!}\\ & \colFR{blue}{$\sqrt{a}\sqrt{b} > 0$}\\ \\ よって, & \sqrt{a}\sqrt{b}\ は,\\ & \colFR{red}{2乗して\ $ab$\ になる}\,\colFR{blue}{正の数である}から,\\ \\ & \sqrt{a}\sqrt{b} = \sqrt{ab}\\ & \colMM{black}{\fbox{証明終り}} \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{\fbox{\bf Point} } & \ \colMM{red}{\dfrac{\sqrt{a}}{\sqrt{b}}\ \ \ が\ \ \ \dfrac{a}{b}\ の正の平方根}\\ & \color{red}\dfrac{\sqrt{a}}{\sqrt{b}} = +\sqrt{\dfrac{a}{b}}\\ & \colMM{red}{ であることを示す!}\\ \\ \colMM{black}{\fbox{\bf 証明} } & \colMM{orange}{2乗したら\ \frac{a}{b}\ ➡\ \frac{a}{b}\ の平方根!}\\ \left(\dfrac{\sqrt{a}}{\sqrt{b}}\right)^{\colBX{bisque}{$\scriptsize 2$}} &= \dfrac{(\sqrt{a})^2}{(\sqrt{b})^2} = \colFR{red}{$\dfrac{a}{b}$}\\ \\ そして, & \color{lightgray}a>0,\ b>0\ より\\ \\ & \sqrt{a}>0,\ \sqrt{b}>0\ より\\ & \colMM{green}{ \swarrow\dfrac{\sqrt{a}}{\sqrt{b}}\ は正!}\\ & \colFR{blue}{$\dfrac{\sqrt{a}}{\sqrt{b}} > 0$}\\ \\ よって, & \dfrac{\sqrt{a}}{\sqrt{b}}\ は,\\ & \colFR{red}{2乗して\ $\dfrac{a}{b}$\ になる}\,\colFR{blue}{正の数である}から,\\ \\ & \dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}\\ & \colMM{black}{\fbox{\bf 証明終り}} \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
次の式を計算しよう。
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡素因数分解!}\\ \sqrt{20} &= \sqrt{2 \cdot 2 \times 5}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{2 \cdot 2} \times \sqrt{5}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{2^2} \times \sqrt{5}\\ & \colMM{red}{ \Darr \sqrt{2^2} = |\,2\,| = 2}\\ &= 2\sqrt{5} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡まとめる}\\ \sqrt{6}\sqrt{15} &= \sqrt{6 \times 15}\\ & \colMM{red}{ \Darr 素因数分解!}\\ &= \sqrt{2 \cdot 3 \times 3 \cdot 5}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{3 \cdot 3} \times \sqrt{2 \cdot 5}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{3^2} \times \sqrt{10}\\ & \colMM{red}{ \Darr \sqrt{3^2} = |\,3\,| = 3}\\ &= 3\sqrt{10} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡まとめる}\\ \dfrac{\sqrt{84}}{\sqrt{3}} &= \sqrt{\dfrac{84}{3}}\\ & \colMM{red}{ \Darr 約分}\\ &= \sqrt{28}\\ & \colMM{red}{ \Darr 素因数分解!}\\ &= \sqrt{2 \cdot 2 \cdot 7}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{2 \cdot 2} \times \sqrt{7}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{2^2} \times \sqrt{7}\\ & \colMM{red}{ \Darr \sqrt{2^2} = |\,2\,| = 2}\\ &= 2\sqrt{7} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡素因数分解!}\\ \sqrt{125} &= \sqrt{5 \cdot 5 \cdot 5}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{5 \cdot 5} \times \sqrt{5}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{5^2} \times \sqrt{5}\\ & \colMM{red}{ \Darr \sqrt{5^2} = |\,5\,| = 5}\\ &= 5\sqrt{5} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡まとめる}\\ \sqrt{6}\sqrt{8} &= \sqrt{6 \times 8}\\ & \colMM{red}{ \Darr 素因数分解!}\\ &= \sqrt{2 \cdot 3 \times 2 \cdot 2 \cdot 2}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{2 \cdot 2} \times \sqrt{2 \cdot 2} \times \sqrt{3}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{2^2} \times \sqrt{2^2} \times \sqrt{3}\\ & \colMM{red}{ \Darr \sqrt{2^2} = |\,2\,| = 2}\\ &= 2 \times 2 \times \sqrt{3}\\ \\ &= 4\sqrt{3} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡まとめる}\\ \dfrac{\sqrt{108}}{\sqrt{2}} &= \sqrt{\dfrac{108}{2}}\\ & \colMM{red}{ \Darr 約分}\\ &= \sqrt{54} \color{lightgray}=\sqrt{6 \times 9}\\ & \colMM{red}{ \Darr 素因数分解!}\\ &= \sqrt{2 \cdot 3 \cdot 3 \cdot 3}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{3 \cdot 3} \times \sqrt{2 \cdot 3}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{3^2} \times \sqrt{6}\\ & \colMM{red}{ \Darr \sqrt{3^2} = |\,3\,| = 3}\\ &= 3\sqrt{6} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡素因数分解!}\\ \sqrt{81} &= \sqrt{9 \cdot 9}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{3 \cdot 3} \times \sqrt{3 \cdot 3}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{3^2} \times \sqrt{3^2}\\ & \colMM{red}{ \Darr \sqrt{3^2} = |\,3\,| = 3}\\ &= 3 \cdot 3 = 9 \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡まとめる}\\ \sqrt{10}\sqrt{5} &= \sqrt{10 \times 5}\\ & \colMM{red}{ \Darr 素因数分解!}\\ &= \sqrt{5 \cdot 2 \times 5}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{5 \cdot 5} \times \sqrt{2}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{5^2} \times \sqrt{2}\\ & \colMM{red}{ \Darr \sqrt{5^2} = |\,5\,| = 5}\\ &= 5\sqrt{2} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{➡まとめる}\\ \dfrac{\sqrt{36}}{\sqrt{3}} &= \sqrt{\dfrac{36}{3}}\\ & \colMM{red}{ \Darr 約分}\\ &= \sqrt{12} \color{lightgray}=\sqrt{3 \times 4}\\ & \colMM{red}{ \Darr 素因数分解!}\\ &= \sqrt{3 \cdot 2 \cdot 2}\\ & \colMM{red}{ \Darr ばらす}\\ &= \sqrt{2 \cdot 2} \times \sqrt{3}\\ & \colMM{red}{ \Darr 2乗発見!}\\ &= \sqrt{2^2} \times \sqrt{3}\\ & \colMM{red}{ \Darr \sqrt{2^2} = |\,2\,| = 2}\\ &= 2\sqrt{3} \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
ルートをはずしてから計算しよう!
↓この問題へのリンクはこちら(右クリックで保存)
次の式を計算しよう。
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \sqrt{3}+\sqrt{48}-\sqrt{27}\\ & \colMM{orange}{ \Darr \sqrt{ルートの中身}をかけ算になおす!}\\ &= \sqrt{3}+\sqrt{6 \times 8}-\sqrt{3 \times 9}\\ & \colMM{orange}{ \Darr 素因数分解!}\\ &= \sqrt{3} +\sqrt{2 \cdot 3 \times 2 \cdot 2 \cdot 2} - \sqrt{3 \times 3 \cdot 3}\\ & \colMM{green}{ \Darr 2つずつ組み合わせて分解}\\ &= \sqrt{3} + \sqrt{2^2}\sqrt{2^2}\sqrt{3}-\sqrt{3^2}\sqrt{3}\\ & \colMM{red}{ \Darr ★\ \sqrt{a^2}=|\,a\,|}\\ &= \sqrt{3}+2\cdot 2\sqrt{3}-3\sqrt{3}\\ \\ &= \sqrt{3}+4\sqrt{3}-3\sqrt{3}\\ & \colMM{magenta}{ \Darr※\ x+4x-3x=2x\ と同じ計算!}\\ &\color{lightgray}= (1+4-3)\sqrt{3}\\ \\ &= 2\sqrt{3} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \sqrt{8}+\sqrt{50}\\ & \colMM{orange}{ \Darr \sqrt{ルートの中身}をかけ算になおす!}\\ &= \sqrt{2 \times 4}+\sqrt{5 \times 10}\\ & \colMM{orange}{ \Darr 素因数分解!}\\ &= \sqrt{2 \times 2 \cdot 2} +\sqrt{5 \times 5 \cdot 2}\\ & \colMM{green}{ \Darr 2つずつ組み合わせて分解}\\ &= \sqrt{2^2}\sqrt{2}+\sqrt{5^2}\sqrt{2}\\ & \colMM{red}{ \Darr ★\ \sqrt{a^2}=|\,a\,|\ ★}\\ &= 2\sqrt{2}+5\sqrt{2}\\ & \colMM{magenta}{ \Darr ※\ 2x+5x=7x\ と同じ計算!}\\ &\color{lightgray}= (2+5)\sqrt{2}\\ \\ &= 7\sqrt{2} \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】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & 3\sqrt{27}+2\sqrt{12}-\sqrt{75}\\ & \colMM{orange}{ \Darr \sqrt{ルートの中身}をかけ算になおす!}\\ &= 3\sqrt{3 \times 9}+2\sqrt{3 \times 4}-\sqrt{3 \times 25}\\ & \colMM{orange}{ \Darr 素因数分解!}\\ &= 3\sqrt{3 \times 3 \cdot 3}+2\sqrt{3 \times 2 \cdot 2}-\sqrt{3 \times 5 \cdot 5}\\ & \colMM{green}{ \Darr 2つずつ組み合わせて分解}\\ &= 3\sqrt{3^2}\sqrt{3}+2\sqrt{2^2}\sqrt{3}-\sqrt{5^2}\sqrt{3}\\ & \colMM{red}{ \Darr ★\ \sqrt{a^2}=|\,a\,|\ ★}\\ &= 9\sqrt{3}+4\sqrt{3}-5\sqrt{3}\\ & \colMM{magenta}{ \Darr ※\ 9x+4x-5x=8x\ と同じ計算!}\\ &\color{lightgray}= (9+4-5)\sqrt{3}\\ \\ &= 8\sqrt{3} \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
展開の公式を利用しよう!
次の式を計算しよう。
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\def\SL{\sqrt{3}} \def\SF{+} \def\SR{\sqrt{5}} \def\SLZ{3} \def\SRZ{5} \def\SLRN{2\sqrt{15}} \def\Kotae{8+2\sqrt{15}} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{\bf \fbox{\bf 準備} 慣れる} & \colMM{red}{\bf までは余白で計算しよう!}\\ & (\colBX{bisque}{$\SL$}\SF\colBX{palegreen}{$\SR$})^2\\ \\ \colMM{red}{左^2 = } & \left(\colBX{bisque}{$\SL$}\right)^2 = \SLZ \\ \colMM{red}{右^2 = } & \left(\colBX{palegreen}{$\SR$}\right)^2 = \SRZ\\ \\ \colMM{red}{両方かけて} & \colMM{red}{2倍}\\ \colMM{red}{左右\times2 =} & \colBX{bisque}{$\SL$} \times \colBX{palegreen}{$\SR$} \times2 = \SLRN\\ \\ \colMM{red}{\bf \fbox{\bf 解答} 準備し} & \colMM{red}{\bf た結果を足すだけ!}\\ (\SL\SF\SR)^2 &= \SLZ\SF\SLRN+\SRZ\\ &= \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\SL{\sqrt{2}} \def\SF{} \def\SR{-\sqrt{3}} \def\SLZ{2} \def\SRZ{3} \def\SLRN{-2\sqrt{6}} \def\Kotae{5-2\sqrt{6}} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} \colMM{red}{\bf \fbox{\bf 準備} 慣れる} & \colMM{red}{\bf までは余白で計算しよう!}\\ & (\colBX{bisque}{$\SL$}\SF\colBX{palegreen}{$\SR$})^2\\ \\ \colMM{red}{左^2 = } & \left(\colBX{bisque}{$\SL$}\right)^2 = \SLZ \\ \colMM{red}{右^2 = } & \left(\colBX{palegreen}{$\SR$}\right)^2 = \SRZ\\ \\ \colMM{red}{両方かけて} & \colMM{red}{2倍}\\ \colMM{red}{左右\times2 =} & \colBX{bisque}{$\SL$} \times (\colBX{palegreen}{$\SR$}) \times2 = \SLRN\\ \\ \colMM{red}{\bf \fbox{\bf 解答} 準備し} & \colMM{red}{\bf た結果を足すだけ!}\\ (\SL\SF\SR)^2 &= \SLZ\SF\SLRN+\SRZ\\ &= \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\SL{\sqrt{7}} \def\SR{\sqrt{2}} \def\SLZ{7} \def\SRZ{2} \def\Kotae{5} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{ \ \Darr 符号が違うだけ \Darr}\\ & (\colBX{bisque}{$\SL$}+\colBX{palegreen}{$\SR$})(\colBX{bisque}{$\SL$}-\colBX{palegreen}{$\SR$})\\ & \colMM{red}{ 左^2-右^2}\\ &= (\colBX{bisque}{$\SL$})^2 - (\colBX{palegreen}{$\SR$})^2\\ \\ &= \SLZ-\SRZ= \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan