2項×2項の展開の基本

ただいま作成中

私の授業で使いながら問題を増やしているため、完成するまでに時間がかかりそうです。少しずつ問題を増やしたり、ポイント解説を付けたりしていきます。無限の彼方で完成する日を、どうぞご期待ください。

Happy Math-ing!

未完成でもよければ、使ってやってください。😃

次の式を展開しよう。

この問題へのリンクはこちら(右クリックで保存)

【解答】

\def\SLA{x}
\def\SLF{+}
\def\SLB{2y}
\def\SRA{x}
\def\SRF{}
\def\SRB{-3y}
\def\SLRA{x^2}
\def\SLRB{-6y^2}
\def\SLARB{-3xy}
\def\SLBRA{2xy}
\def\SM{-xy}
\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}{慣れるまでは余白にメモしよう!}\\
\\
& \colMM{orange}{\bf 左}  \colMM{magenta}{\bf 右}  \colMM{green}{\bf 左}  \colMM{deepskyblue}{\bf 右}\\
& (\colBX{bisque}{$\SLA$}\SLF\colBX{violet}{$\SLB$})(\colBX{palegreen}{$\SRA$}\SRF\colBX{lightcyan}{$\SRB$})\\
& \colMM{orange}{\bf 外}  \colMM{magenta}{\bf 中}  \colMM{green}{\bf 中}  \colMM{deepskyblue}{\bf 外}\\
\\
・ & \colBX{bisque}{左} \times \colBX{palegreen}{左} = (\colBX{bisque}{$\SLA$}) \times (\colBX{palegreen}{$\SRA$}) = \colFR{red}{$\SLRA$}\\
・ & \colBX{violet}{右} \times \colBX{lightcyan}{右} = (\colBX{violet}{$\SLB$}) \times (\colBX{lightcyan}{$\SRB$}) = \colFR{blue}{$\SLRB$}\\
\\
・ & \colBX{bisque}{外} \times \colBX{lightcyan}{外} = (\colBX{bisque}{$\SLA$}) \times (\colBX{lightcyan}{$\SRB$}) = \SLARB\\
・ & \colBX{violet}{中} \times \colBX{palegreen}{中} = (\colBX{violet}{$\SLB$}) \times (\colBX{palegreen}{$\SRA$}) = \SLBRA\\\hline
& \colMM{red}{                足して} \colBX{yellow}{$\SM$}
\\
\\
\colMM{red}{\fbox{Answer}} &\\
& (\SLA\SLF\SLB)(\SRA\SRF\SRB)
= \colFR{red}{$\SLRA$}\,\colBX{yellow}{$\SM$}\,\colFR{blue}{$\SLRB$}
\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\SLA{x}
\def\SLF{}
\def\SLB{-4y}
\def\SRA{x}
\def\SRF{+}
\def\SRB{2y}
\def\SLRA{x^2}
\def\SLRB{-8y^2}
\def\SLARB{2xy}
\def\SLBRA{-4xy}
\def\SM{-2xy}
\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}{慣れるまでは余白にメモしよう!}\\
\\
& \colMM{orange}{\bf 左}  \colMM{magenta}{\bf 右}  \colMM{green}{\bf 左}  \colMM{deepskyblue}{\bf 右}\\
& (\colBX{bisque}{$\SLA$}\SLF\colBX{violet}{$\SLB$})(\colBX{palegreen}{$\SRA$}\SRF\colBX{lightcyan}{$\SRB$})\\
& \colMM{orange}{\bf 外}  \colMM{magenta}{\bf 中}  \colMM{green}{\bf 中}  \colMM{deepskyblue}{\bf 外}\\
\\
・ & \colBX{bisque}{左} \times \colBX{palegreen}{左} = (\colBX{bisque}{$\SLA$}) \times (\colBX{palegreen}{$\SRA$}) = \colFR{red}{$\SLRA$}\\
・ & \colBX{violet}{右} \times \colBX{lightcyan}{右} = (\colBX{violet}{$\SLB$}) \times (\colBX{lightcyan}{$\SRB$}) = \colFR{blue}{$\SLRB$}\\
\\
・ & \colBX{bisque}{外} \times \colBX{lightcyan}{外} = (\colBX{bisque}{$\SLA$}) \times (\colBX{lightcyan}{$\SRB$}) = \SLARB\\
・ & \colBX{violet}{中} \times \colBX{palegreen}{中} = (\colBX{violet}{$\SLB$}) \times (\colBX{palegreen}{$\SRA$}) = \SLBRA\\\hline
& \colMM{red}{                足して} \colBX{yellow}{$\SM$}
\\
\\
\colMM{red}{\fbox{Answer}} &\\
& (\SLA\SLF\SLB)(\SRA\SRF\SRB)
= \colFR{red}{$\SLRA$}\,\colBX{yellow}{$\SM$}\,\colFR{blue}{$\SLRB$}
\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\SLA{2x}
\def\SLF{+}
\def\SLB{3}
\def\SRA{4x}
\def\SRF{+}
\def\SRB{5}
\def\SLRA{8x^2}
\def\SLRBF{+}
\def\SLRB{15}
\def\SLARB{10x}
\def\SLBRA{12x}
\def\SMF{+}
\def\SM{22x}
\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}{慣れるまでは余白にメモしよう!}\\
\\
& \colMM{orange}{\bf 左}  \colMM{magenta}{\bf 右}  \colMM{green}{\bf 左}  \colMM{deepskyblue}{\bf 右}\\
& (\colBX{bisque}{$\SLA$}\SLF\colBX{violet}{$\SLB$})(\colBX{palegreen}{$\SRA$}\SRF\colBX{lightcyan}{$\SRB$})\\
& \colMM{orange}{\bf 外}  \colMM{magenta}{\bf 中}  \colMM{green}{\bf 中}  \colMM{deepskyblue}{\bf 外}\\
\\
・ & \colBX{bisque}{左} \times \colBX{palegreen}{左} = (\colBX{bisque}{$\SLA$}) \times (\colBX{palegreen}{$\SRA$}) = \colFR{red}{$\SLRA$}\\
・ & \colBX{violet}{右} \times \colBX{lightcyan}{右} = (\colBX{violet}{$\SLB$}) \times (\colBX{lightcyan}{$\SRB$}) = \colFR{blue}{$\SLRB$}\\
\\
・ & \colBX{bisque}{外} \times \colBX{lightcyan}{外} = (\colBX{bisque}{$\SLA$}) \times (\colBX{lightcyan}{$\SRB$}) = \SLARB\\
・ & \colBX{violet}{中} \times \colBX{palegreen}{中} = (\colBX{violet}{$\SLB$}) \times (\colBX{palegreen}{$\SRA$}) = \SLBRA\\\hline
& \colMM{red}{                足して} \colBX{yellow}{$\SM$}
\\
\\
\colMM{red}{\fbox{Answer}} &\\
& (\SLA\SLF\SLB)(\SRA\SRF\SRB)
= \colFR{red}{$\SLRA$}\,\colBX{yellow}{$\SMF\SM$}\,\colFR{blue}{$\SLRBF\SLRB$}
\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\SLA{3x}
\def\SLF{+}
\def\SLB{1}
\def\SRA{x}
\def\SRF{+}
\def\SRB{2}
\def\SLRA{3x^2}
\def\SLRBF{+}
\def\SLRB{2}
\def\SLARB{6x}
\def\SLBRA{x}
\def\SMF{+}
\def\SM{7x}
\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}{慣れるまでは余白にメモしよう!}\\
\\
& \colMM{orange}{\bf 左}  \colMM{magenta}{\bf 右}  \colMM{green}{\bf 左}  \colMM{deepskyblue}{\bf 右}\\
& (\colBX{bisque}{$\SLA$}\SLF\colBX{violet}{$\SLB$})(\colBX{palegreen}{$\SRA$}\SRF\colBX{lightcyan}{$\SRB$})\\
& \colMM{orange}{\bf 外}  \colMM{magenta}{\bf 中}  \colMM{green}{\bf 中}  \colMM{deepskyblue}{\bf 外}\\
\\
・ & \colBX{bisque}{左} \times \colBX{palegreen}{左} = (\colBX{bisque}{$\SLA$}) \times (\colBX{palegreen}{$\SRA$}) = \colFR{red}{$\SLRA$}\\
・ & \colBX{violet}{右} \times \colBX{lightcyan}{右} = (\colBX{violet}{$\SLB$}) \times (\colBX{lightcyan}{$\SRB$}) = \colFR{blue}{$\SLRB$}\\
\\
・ & \colBX{bisque}{外} \times \colBX{lightcyan}{外} = (\colBX{bisque}{$\SLA$}) \times (\colBX{lightcyan}{$\SRB$}) = \SLARB\\
・ & \colBX{violet}{中} \times \colBX{palegreen}{中} = (\colBX{violet}{$\SLB$}) \times (\colBX{palegreen}{$\SRA$}) = \SLBRA\\\hline
& \colMM{red}{                足して} \colBX{yellow}{$\SM$}
\\
\\
\colMM{red}{\fbox{Answer}} &\\
& (\SLA\SLF\SLB)(\SRA\SRF\SRB)
= \colFR{red}{$\SLRA$}\,\colBX{yellow}{$\SMF\SM$}\,\colFR{blue}{$\SLRBF\SLRB$}
\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\SLA{x}
\def\SLF{+}
\def\SLB{2}
\def\SRA{x}
\def\SRF{+}
\def\SRB{7}
\def\SLRA{x^2}
\def\SLRBF{+}
\def\SLRB{14}
\def\SLARB{7x}
\def\SLBRA{2x}
\def\SMF{+}
\def\SM{9x}
\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}{慣れるまでは余白にメモしよう!}\\
\\
& \colMM{orange}{\bf 左}  \colMM{magenta}{\bf 右}  \colMM{green}{\bf 左}  \colMM{deepskyblue}{\bf 右}\\
& (\colBX{bisque}{$\SLA$}\SLF\colBX{violet}{$\SLB$})(\colBX{palegreen}{$\SRA$}\SRF\colBX{lightcyan}{$\SRB$})\\
& \colMM{orange}{\bf 外}  \colMM{magenta}{\bf 中}  \colMM{green}{\bf 中}  \colMM{deepskyblue}{\bf 外}\\
\\
・ & \colBX{bisque}{左} \times \colBX{palegreen}{左} = (\colBX{bisque}{$\SLA$}) \times (\colBX{palegreen}{$\SRA$}) = \colFR{red}{$\SLRA$}\\
・ & \colBX{violet}{右} \times \colBX{lightcyan}{右} = (\colBX{violet}{$\SLB$}) \times (\colBX{lightcyan}{$\SRB$}) = \colFR{blue}{$\SLRB$}\\
\\
・ & \colBX{bisque}{外} \times \colBX{lightcyan}{外} = (\colBX{bisque}{$\SLA$}) \times (\colBX{lightcyan}{$\SRB$}) = \SLARB\\
・ & \colBX{violet}{中} \times \colBX{palegreen}{中} = (\colBX{violet}{$\SLB$}) \times (\colBX{palegreen}{$\SRA$}) = \SLBRA\\\hline
& \colMM{red}{                足して} \colBX{yellow}{$\SM$}
\\
\\
\colMM{red}{\fbox{Answer}} &\\
& (\SLA\SLF\SLB)(\SRA\SRF\SRB)
= \colFR{red}{$\SLRA$}\,\colBX{yellow}{$\SMF\SM$}\,\colFR{blue}{$\SLRBF\SLRB$}
\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\SLA{2x}
\def\SLF{+}
\def\SLB{1}
\def\SRA{x}
\def\SRF{+}
\def\SRB{2}
\def\SLRA{2x^2}
\def\SLRBF{+}
\def\SLRB{2}
\def\SLARB{4x}
\def\SLBRA{x}
\def\SMF{+}
\def\SM{5x}
\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}{慣れるまでは余白にメモしよう!}\\
\\
& \colMM{orange}{\bf 左}  \colMM{magenta}{\bf 右}  \colMM{green}{\bf 左}  \colMM{deepskyblue}{\bf 右}\\
& (\colBX{bisque}{$\SLA$}\SLF\colBX{violet}{$\SLB$})(\colBX{palegreen}{$\SRA$}\SRF\colBX{lightcyan}{$\SRB$})\\
& \colMM{orange}{\bf 外}  \colMM{magenta}{\bf 中}  \colMM{green}{\bf 中}  \colMM{deepskyblue}{\bf 外}\\
\\
・ & \colBX{bisque}{左} \times \colBX{palegreen}{左} = (\colBX{bisque}{$\SLA$}) \times (\colBX{palegreen}{$\SRA$}) = \colFR{red}{$\SLRA$}\\
・ & \colBX{violet}{右} \times \colBX{lightcyan}{右} = (\colBX{violet}{$\SLB$}) \times (\colBX{lightcyan}{$\SRB$}) = \colFR{blue}{$\SLRB$}\\
\\
・ & \colBX{bisque}{外} \times \colBX{lightcyan}{外} = (\colBX{bisque}{$\SLA$}) \times (\colBX{lightcyan}{$\SRB$}) = \SLARB\\
・ & \colBX{violet}{中} \times \colBX{palegreen}{中} = (\colBX{violet}{$\SLB$}) \times (\colBX{palegreen}{$\SRA$}) = \SLBRA\\\hline
& \colMM{red}{                足して} \colBX{yellow}{$\SM$}
\\
\\
\colMM{red}{\fbox{Answer}} &\\
& (\SLA\SLF\SLB)(\SRA\SRF\SRB)
= \colFR{red}{$\SLRA$}\,\colBX{yellow}{$\SMF\SM$}\,\colFR{blue}{$\SLRBF\SLRB$}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

レベルG

次の計算をせよ。

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]{\textcolor{#1}{#2}}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& (3x+1)(x-2)-x(2-x)\\
&= 3x(x-2)+1(x-2)-x(2-x)\\
&= 3x^2-6x+x-2-2x+x^2\\
&= 4x^2-7x-2
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【別解】

\newcommand\colNS[2]{\textcolor{#1}{#2}}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& (3x+1)(x-2)-x(2-x)\\
&= (3x+1)(x-2)+x(x-2)\\
&= (3x+1+x)(x-2)\\
&= (4x+1)(x-2)\\
&= 4x(x-2)+1(x-2)\\
&= 4x^2-8x+x-2\\
&= 4x^2-7x-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]{\textcolor{#1}{#2}}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& (3a-1)(a+2)-5a(a+2)\\
&= 3a(a+2)-1(a+2)-5a(a+2)\\
&= 3a^2+6a-a-2-5a^2-10a\\
&= -2a^2-5a-2
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【別解】

\newcommand\colNS[2]{\textcolor{#1}{#2}}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& (3a-1)(a+2)-5a(a+2)\\
&= (3a-1-5a)(a+2)\\
&= (-2a-1)(a+2)\\
&= -2a(a+2)-1(a+2)\\
&= -2a^2-4a-a-2\\
&= -2a^2-5a-2
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

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