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\SL{2x}
\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*}
& \colBX{bisque}{$\SL$}(x^2+2x-3)\\
\\
&= \colBX{bisque}{$\SL$} \cdot x^2 + \colBX{bisque}{$\SL$} \cdot 2x + \colBX{bisque}{$\SL$} \cdot (-3)\\
\\
&= 2x^3+4x^2-6x
\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{x^2}
\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*}
& \colBX{bisque}{$\SL$}(x^2-3x+4)\\
\\
&= \colBX{bisque}{$\SL$} \cdot x^2 + \colBX{bisque}{$\SL$} \cdot (-3x) + \colBX{bisque}{$\SL$} \cdot 4\\
\\
&= x^4-3x^3+4x^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】
\def\SL{2x^2}
\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*}
& \colBX{bisque}{$\SL$}(x^2+2x-1)\\
\\
&= \colBX{bisque}{$\SL$} \cdot x^2 + \colBX{bisque}{$\SL$} \cdot 2x + \colBX{bisque}{$\SL$} \cdot (-1)\\
\\
&= 2x^4+4x^3-2x^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】
\def\SR{xy^2}
\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*}
& (x^3-6y)\colBX{bisque}{$\SR$}\\
\\
&= x^3 \cdot \colBX{bisque}{$\SR$} -6y \cdot \colBX{bisque}{$\SR$}\\
\\
&= x^4y^2-6xy^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】
\def\SR{x^2y}
\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*}
& (xy+3)\colBX{bisque}{$\SR$}\\
\\
&= xy \cdot \colBX{bisque}{$\SR$} +3 \cdot \colBX{bisque}{$\SR$}\\
\\
&= x^3y^2+3x^2y
\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\SR{ab^2}
\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*}
& (a^2-ab)\colBX{bisque}{$\SR$}\\
\\
&= a^2 \cdot \colBX{bisque}{$\SR$} -ab \cdot \colBX{bisque}{$\SR$}\\
\\
&= a^3b^2-a^2b^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】前から
\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*}
& (\colBX{bisque}{$2x$}\,\colBX{palegreen}{$-3$})(3x^2-x+1)\\
\\
&= \colBX{bisque}{$2x$}(3x^2-x+1)\colBX{palegreen}{$-3$}(3x^2-x+1)\\
\\
&= (6x^3-2x^2+2x)+(-9x^2+3x-3)\\
\\
&= 6x^3-2x^2+2x-9x^2+3x-3\\
\\
&\color{lightgray}= 6x^3+(-2x^2-9x^2)+(2x+3x)-3\\
\\
&= 6x^3-11x^2+5x-3
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan【解答】後ろから
\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*}
& (2x-3)(\colBX{bisque}{$3x^2$}\,\colBX{palegreen}{$-x$}\,\colBX{violet}{$+1$})\\
\\
&= (2x-3) \colBX{bisque}{$3x^2$}+(2x-3)\colBX{palegreen}{$(-x)$}+(2x-3)\colBX{violet}{$(+1)$}\\
\\
&= (6x^3-9x^2)+(-2x^2+3x)+(2x-3)\\
\\
&= 6x^3-9x^2-2x^2+3x+2x-3\\
\\
&\color{lightgray}= 6x^3+(-9x^2-2x^2)+(3x+2x)-3\\
\\
&= 6x^3-11x^2+5x-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】前から
\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*}
& (\colBX{bisque}{$3x$}\,\colBX{palegreen}{$-1$})(2x^2-x+5)\\
\\
&= \colBX{bisque}{$3x$}(2x^2-x+5)\colBX{palegreen}{$-1$}(2x^2-x+5)\\
\\
&= (6x^3-3x^2+15x)+(-2x^2+x-5)\\
\\
&= 6x^3-3x^2+15x-2x^2+x-5\\
\\
&\color{lightgray}= 6x^3+(-3x^2-2x^2)+(15x+x)-5\\
\\
&= 6x^3-5x^2+16x-5
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan【解答】後ろから
\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*}
& (3x-1)(\colBX{bisque}{$2x^2$}\,\colBX{palegreen}{$-x$}\,\colBX{violet}{$+5$})\\
\\
&= (3x-1) \colBX{bisque}{$2x^2$}+(3x-1)\colBX{palegreen}{$(-x)$}+(3x-1)\colBX{violet}{$(+5)$}\\
\\
&= (6x^3-2x^2)+(-3x^2+x)+(15x-5)\\
\\
&= 6x^3-2x^2-3x^2+x+15x-5\\
\\
&\color{lightgray}= 6x^3+(-2x^2-3x^2)+(x+15x)-5\\
\\
&= 6x^3-5x^2+16x-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】前から
\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*}
& (\colBX{bisque}{$3x$}\,\colBX{palegreen}{$-1$})(x^2-2x+3)\\
\\
&= \colBX{bisque}{$3x$}(x^2-2x+3)\colBX{palegreen}{$-1$}(x^2-2x+3)\\
\\
&= (3x^3-6x^2+9x)+(-x^2+2x-3)\\
\\
&= 3x^3-6x^2+9x-x^2+2x-3\\
\\
&\color{lightgray}= 3x^3+(-6x^2-x^2)+(9x+2x)-3\\
\\
&= 3x^3-7x^2+11x-3
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan【解答】後ろから
\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*}
& (3x-1)(\colBX{bisque}{$x^2$}\,\colBX{palegreen}{$-2x$}\,\colBX{violet}{$+3$})\\
\\
&= (3x-1) \colBX{bisque}{$x^2$}+(3x-1)\colBX{palegreen}{$(-2x)$}+(3x-1)\colBX{violet}{$(+3)$}\\
\\
&= (3x^3-x^2)+(-6x^2+2x)+(9x-3)\\
\\
&= 3x^3-x^2-6x^2+2x+9x-3\\
\\
&\color{lightgray}= 3x^3+(-x^2-6x^2)+(2x+9x)-3\\
\\
&= 3x^3-7x^2+11x-3
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan