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