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３次方程式を解け。
↓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-1}
\def\sr{x^2+x+1}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= 1
}
\def\sright{%
x &= \dfrac{-1\pm\sqrt{1-4 \cdot 1}}{2}\\
&= \dfrac{-1\pm\sqrt{-3}}{2}\\
&= \dfrac{-1\pm\sqrt{3}\,i}{2}
}
\def\kotae{1,\ \dfrac{-1\pm\sqrt{3}\,i}{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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって\\
& x = \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3-1}
\def\ls{x}
\def\fg{-}
\def\rs{1}
\def\ms{+x}
\def\lrs{-x}
\def\kright{x^2+x+1}
\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*}
& \colMM{red}{\bf 3乗\fg3乗\ に変形する}\\
\siki &= \ls^3 \fg \rs^3\\
\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\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}
\def\sr{x^2+2x+4}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= 2
}
\def\sright{%
x &= \dfrac{-2\pm\sqrt{4-4 \cdot 4}}{2}\\
&= \dfrac{-2\pm\sqrt{-12}}{2}\\
&= \dfrac{-2\pm\sqrt{12}\,i}{2}\\
&= \dfrac{-2\pm2\sqrt{3}\,i}{2}\\
&= \dfrac{2(-1\pm\sqrt{3}\,i)}{2} = -1\pm\sqrt{3}\,i\\
}
\def\kotae{2,\ -1\pm\sqrt{3}\,i}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって\\
& x = \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3-8}
\def\ls{x}
\def\fg{-}
\def\rs{2}
\def\ms{+2x}
\def\lrs{-2x}
\def\kright{x^2+2x+4}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\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+1}
\def\sr{x^2-x+1}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= -1
}
\def\sright{%
x &= \dfrac{1\pm\sqrt{1-4 \cdot 1}}{2}\\
&= \dfrac{1\pm\sqrt{-3}}{2}\\
&= \dfrac{1\pm\sqrt{3}\,i}{2}
}
\def\kotae{-1,\ \dfrac{1\pm\sqrt{3}\,i}{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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって\\
& x = \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3+1}
\def\ls{x}
\def\fg{+}
\def\rs{1}
\def\ms{-x}
\def\lrs{+x}
\def\kright{x^2-x+1}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\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-3}
\def\sr{x^2+3x+9}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= 3
}
\def\sright{%
x &= \dfrac{-3\pm\sqrt{9-4 \cdot 9}}{2}\\
&= \dfrac{-3\pm\sqrt{-27}}{2}\\
&= \dfrac{-3\pm\sqrt{27}\,i}{2}\\
&= \dfrac{-3\pm3\sqrt{3}\,i}{2}
}
\def\kotae{3,\ \dfrac{-3\pm3\sqrt{3}\,i}{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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって\\
& x = \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3-27}
\def\ls{x}
\def\fg{-}
\def\rs{3}
\def\lrs{-3x}
\def\ms{+3x}
\def\kright{x^2+3x+9}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\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}
\def\sr{x^2-2x+4}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= -2
}
\def\sright{%
x &= \dfrac{2\pm\sqrt{4-4 \cdot 4}}{2}\\
&= \dfrac{2\pm\sqrt{-12}}{2}\\
&= \dfrac{2\pm\sqrt{12}\,i}{2}\\
&= \dfrac{2\pm2\sqrt{3}\,i}{2}\\
&= \dfrac{2(1\pm\sqrt{3}\,i)}{2}=1\pm\sqrt{3}\,i
}
\def\kotae{-2,\ 1\pm\sqrt{3}\,i}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって\\
& x = \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3+8}
\def\ls{x}
\def\fg{+}
\def\rs{2}
\def\lrs{+2x}
\def\ms{-2x}
\def\kright{x^2-2x+4}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\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-4}
\def\sr{x^2+4x+16}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= 4
}
\def\sright{%
x &= \dfrac{-4\pm\sqrt{16-4 \cdot 16}}{2}\\
&= \dfrac{-4\pm\sqrt{-48}}{2}\\
&= \dfrac{-4\pm\sqrt{48}\,i}{2}\\
&= \dfrac{-4\pm4\sqrt{3}\,i}{2}\\
&= \dfrac{4(-1\pm\sqrt{3}\,i)}{2}\\
& =2(-1\pm\sqrt{3}\,i) = -2\pm2\sqrt{3}\,i
}
\def\kotae{4,\ -2\pm2\sqrt{3}\,i}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& x^3=64\ より\ x^3-64=0\\
\\
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって\\
& x = \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3-64}
\def\ls{x}
\def\fg{-}
\def\rs{4}
\def\lrs{-4x}
\def\ms{+4x}
\def\kright{x^2+4x+16}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan次のものを求めよ。
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\mv{27}
\def\sl{x-3}
\def\sr{x^2+3x+9}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= 3
}
\def\sright{%
x &= \dfrac{-3\pm\sqrt{9-4 \cdot 9}}{2}\\
&= \dfrac{-3\pm\sqrt{-27}}{2}\\
&= \dfrac{-3\pm\sqrt{27}\,i}{2}\\
&= \dfrac{-3\pm3\sqrt{3}\,i}{2}
}
\def\kotae{3,\ \dfrac{-3\pm3\sqrt{3}\,i}{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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& \mv\ の平方根を\ x\ とすると\\
& x^3=\mv\ より\ x^3-\mv=0\\
\\
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって，\mv\ の３乗根は\\
& \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3-27}
\def\ls{x}
\def\fg{-}
\def\rs{3}
\def\lrs{-3x}
\def\ms{+3x}
\def\kright{x^2+3x+9}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\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\mv{-8}
\def\henkei{x^3=-8\ より\ x^3+8=0}
\def\sl{x+2}
\def\sr{x^2-2x+4}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= -2
}
\def\sright{%
x &= \dfrac{2\pm\sqrt{4-4 \cdot 4}}{2}\\
&= \dfrac{2\pm\sqrt{-12}}{2}\\
&= \dfrac{2\pm\sqrt{12}\,i}{2}\\
&= \dfrac{2\pm2\sqrt{3}\,i}{2}\\
&= \dfrac{2(1\pm\sqrt{3}\,i)}{2} = 1\pm\sqrt{3}\,i\\
}
\def\kotae{-2,\ 1\pm\sqrt{3}\,i}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& \mv\ の平方根を\ x\ とすると\\
& \henkei\\
\\
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって，\mv\ の３乗根は\\
& \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3+8}
\def\ls{x}
\def\fg{+}
\def\rs{2}
\def\lrs{+2x}
\def\ms{-2x}
\def\kright{x^2-2x+4}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\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\mv{8}
\def\henkei{x^3=8\ より\ x^3-8=0}
\def\sl{x-2}
\def\sr{x^2+2x+4}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= 2
}
\def\sright{%
x &= \dfrac{-2\pm\sqrt{4-4 \cdot 4}}{2}\\
&= \dfrac{-2\pm\sqrt{-12}}{2}\\
&= \dfrac{-2\pm\sqrt{12}\,i}{2}\\
&= \dfrac{-2\pm2\sqrt{3}\,i}{2}\\
&= \dfrac{2(-1\pm\sqrt{3}\,i)}{2} = -1\pm\sqrt{3}\,i\\
}
\def\kotae{2,\ -1\pm\sqrt{3}\,i}
\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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& \mv\ の平方根を\ x\ とすると\\
& \henkei\\
\\
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって，\mv\ の３乗根は\\
& \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3-8}
\def\ls{x}
\def\fg{-}
\def\rs{2}
\def\lrs{-2x}
\def\ms{+2x}
\def\kright{x^2+2x+4}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\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\mv{1}
\def\henkei{x^3=1\ より\ x^3-1=0}
\def\sl{x-1}
\def\sr{x^2+x+1}
\def\insubunkai{%
(\sl)(\sr) &= 0\\
}
\def\bunkai{\sl=0{\scriptsize または }\sr=0}
\def\sleft{%
\sl &= 0\\
x &= 1
}
\def\sright{%
x &= \dfrac{-1\pm\sqrt{1-4 \cdot 1}}{2}\\
&= \dfrac{-1\pm\sqrt{-3}}{2}\\
&= \dfrac{-1\pm\sqrt{3}\,i}{2}
}
\def\kotae{1,\ \dfrac{-1\pm\sqrt{3}\,i}{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}}}
\newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}}
\begin{align*}
& \mv\ の平方根を\ x\ とすると\\
& \henkei\\
\\
& 左辺を因数分解すると\colMM{orange}{\bf ★下に解説}\\
& \begin{align*}
\insubunkai
\end{align*}\\
\\
& よって\\
& \bunkai\\
\\
& \begin{align*}
\sleft
\end{align*}\\
\\
& \sr＝０\\
& \begin{align*}
\sright
\end{align*}\\
\\
& したがって，\mv\ の３乗根は\\
& \kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan★左辺を因数分解すると
\def\siki{x^3-1}
\def\ls{x}
\def\fg{-}
\def\rs{1}
\def\lrs{-x}
\def\ms{+x}
\def\kright{x^2+x+1}
\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*}
\siki &= \ls^3 \fg \rs^3\\
& \colMM{orange}{\bf \Darr ３乗とる}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{white}{\ls^2\ms+\rs^2})}\\
& \colMM{green}{\bf \ \ 左^2 右^2}\\
&\colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{white}{\ms}\colNS{green}{+\rs^2})}\\
& \colMM{magenta}{ \ \ {\bf 左} \times {\bf 右} = \lrs\ \Darr\ \bf符号をかえる}\\
& \colNS{orange}{=(\ls\fg\rs)(\colNS{green}{\ls^2}\colNS{magenta}{\ms}\colNS{green}{+\rs^2})}\\
\\
& = (\ls\fg\rs)(\kright)
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan