%
%	gbb60166@gmail.com	https://unilab.gbb60166.jp/prekou/prekou.htm
%
% aspectratio= は 1610, 169, 149, 54, 32 の中から選べる（省略時は 43）
% C:\texlive\2024\texmf-dist\tex\latex\beamer\beamer.cls
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% テーマ
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% navi. symbolsは目立たないが，dvipdfmxを使うと機能しないので非表示に
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%\usepackage{graphicx}
%\usepackage{tkokugo,furikana,tsayusen,shiika,sfkanbun,jdkintou,plext}
\usepackage{type1cm}
\usepackage[deluxe]{otf}
%\usepackage{tikzsymbols}
%\usepackage[dvipdfmx]{graphicx}%
\def\pgfsysdriver{pgfsys-dvipdfmx.def}%（graphicxパッケージを使用しない場合はこの行を有効に）
%\def\pgfsysdriver{pgfsys-dvips.def}%デフォルト
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\usepackage{tikz}%（これで、pgfとpgfforが読み込まれます。）
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\hypersetup{% Jul 18 2023
	pdftitle={プレ高数学科（LaTeX Beamer で汎用PDFプレゼンを作る）},
	pdfauthor={gbb60166@gmail.com},
	pdfsubject={https://unilab.gbb60166.jp/prekou/prekou.htm}
	}

% フォントはお好みで
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\setbeamertemplate{frametitle}[default][center]
\usefonttheme{professionalfonts}

% 参考にしたURL
% http://windom.phys.hirosaki-u.ac.jp/fswiki/wiki.cgi?page=LaTeX+Beamer%A4%C7%A5%D7%A5%EC%A5%BC%A5%F3%A5%C6%A1%BC%A5%B7%A5%E7%A5%F3

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\everymath{\displaystyle}

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\makeatletter
\newcommand{\hooklen}[2]{\settowidth{\@tempdima}{\(#1\)}
%\advance\@tempdima by.3ex
% ↑ 数式モードで式の前後に入るスペースを制御したかったが、
%    難しいのでやめた。段々難解なコードになっているのでやめた方がよい？
\hbox to\@tempdima{\hfil \(#2\)\hfil}}
%■等号付き不等号 by 美文書作成入門
\newcommand{\myleqq}{\mathrel{\mathpalette\gl@align<}}
\newcommand{\mygeqq}{\mathrel{\mathpalette\gl@align>}}
\newcommand{\gl@align}[2]{\lower.6ex\vbox{\baselineskip\z@skip\lineskip\z@
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% カラーユニバーサルデザインを調べたつもりだがあまり自信がありません
% http://www.fukushihoken.metro.tokyo.jp/kiban/machizukuri/kanren/color.files/colorudguideline.pdf
% http://jfly.iam.u-tokyo.ac.jp/colorset/

% ■ アクセントカラー小面積を目立たせる高彩度色
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\definecolor{CUDGreen}{RGB}{3,175,122}
\definecolor{CUDBlue}{RGB}{0,90,255}
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\definecolor{CUDYellow}{RGB}{255,241,0}
\definecolor{CUDBrown}{RGB}{128,64,0}

\definecolor{CUDOrange}{RGB}{246,170,0}

% ■ ベースカラー広い面積の塗り分けに用いる低・中彩度色
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\definecolor{CUDBrightGreen}{RGB}{119,217,168}
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\newcommand{\myfrac}[2]{\mathchoice%■ 修正 Oct 8,2023
	{%\fontsize{20}{20}\selectfont%	\displaystyle
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	{\textstyle{\frac{#1}{#2}}}%	\textstyle
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	{%\fontsize{11}{11}\selectfont%	\scriptscriptstyle
		\frac{\raisebox{0ex}{\(#1\)}}{\raisebox{-.4ex}{\(#2\)}}}%
	}


\begin{document}

\title{プレ高数学科}\author{gbb60166}

%■ 本文と数式モードの間に挿入される空白スペース
%\mathsurround=-1pt
%■ \( \log\log \)でlogとlogの間に挿入される細スペース（省略形は \,）
%\thinmuskip=3mu
%■ \( a+b \)で+の前後に挿入される中スペース（省略形は \>）
\medmuskip=2mu plus 1mu minus 1mu
%■ \( x=5 \)で=の前後に挿入される太スペース（省略形は \;）
%\thickmuskip=5mu plus 5mu
% mu は math units の省略で長さの単位

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	{\fontsize{7pt}{7pt}\selectfont
		\shortstack{一旦\\[-2pt]停止}};
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	}
%　斜線（１文字を消す）x
\newcommand{\sshasen}{\kern-0.6em\tikz[baseline=2pt]
	\draw[line width=1.6pt, CUDRed](0,0)--(0.5,0.5);}
%　斜線（文字式 x+2 などを消す）
\newcommand{\ssshasen}{\kern-1em\tikz[baseline=2pt]
	\draw[line width=1.6pt, CUDRed](0,0)--(0.8,0.7);}

%■■■■■■■■■■■■■　テスト領域　■■■■■■■■■■■■■■


%\end{document}
%■■■■■■■■■■■■■　完成品　■■■■■■■■■■■■■■

\begin{frame}[t]
\frametitle{\( x=\tfrac{\,3-\sqrt{5}\,}{2} \)
	%分母の数字が切れるので見えない支柱を入れる
	\rule[-1ex]{0pt}{0.2ex}	のとき、次の式の値を求めなさい}
\makebox[0.32\textwidth][l]{\ajKakko{1}\ \( x+\myfrac{1}{\,x\,} \)}
\makebox[0.32\textwidth][l]{\ajKakko{2}\ \( x^2+\myfrac{1}{\,x^2} \)}
\makebox[0.32\textwidth][l]{\ajKakko{3}\ \( x^2-\myfrac{1}{\,x^2} \)}

\medskip
\makebox[0.32\textwidth][l]{\ajKakko{4}\ \( x^4+\myfrac{1}{\,x^4} \)}
\makebox[0.32\textwidth][l]{\ajKakko{5}\ \( x^4-\myfrac{1}{\,x^4} \)}

\bigskip
\colorbox{CUDCream}{対称式}、\colorbox{CUDCream}{交代式}の問題です。

\medskip
まず事前準備として\( \myfrac{1}{\,x\,} \)を計算します。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\setbeamercolor{frametitle}{bg=CUDCream!75,fg=black!80}
\setbeamercolor{palette primary}{bg=CUDCream!75,fg=black!80}

\begin{frame}[t]
\frametitle{\( x=\tfrac{\,3-\sqrt{5}\,}{2} \)
	%分母の数字が切れるので見えない支柱を入れる
	\rule[-1ex]{0pt}{0.2ex}	\quad 事前準備の計算中}
\vskip-5ex
\begin{eqnarray*}
\myfrac{1}{\,x\,} &=& \myfrac{1\textcolor{CUDRed}{\scriptstyle\times2}}
	{\ \tfrac{3-\sqrt{5}}{2}\textcolor{CUDRed}{\scriptstyle\times2}\ }
	\quad = \myfrac{2}{3-\sqrt{5}}
	\quad\text{\footnotesize\textcolor{CUDBlue}{有理化する}}\\
&=& \myfrac{2\,\textcolor{CUDBlue}{(3+\sqrt{5})}}
	{(3-\sqrt{5})\textcolor{CUDBlue}{(3+\sqrt{5})}}
	\quad = \myfrac{2\,(3+\sqrt{5})}{3^2-\sqrt{5}^{\kern0.1em 2}}\\
&=& \myfrac{2\,(3+\sqrt{5})}{9-5}
	= \myfrac{2\sshasen\,(3+\sqrt{5})}{4\sshasen{}_{\textcolor{CUDRed}{2}}}
	= \myfrac{3+\sqrt{5}}{2}\ \ittan
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\setbeamercolor{frametitle}{use=structure,fg=white,bg=structure.fg}
\setbeamercolor{palette primary}{use=structure,fg=white,bg=structure.fg}

\begin{frame}[t]
\frametitle{\( x=\tfrac{\,3-\sqrt{5}\,}{2},\ 
	\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}
	=\tfrac{\,3+\sqrt{5}\,}{2} \)\quad だよ
	\rule[-1ex]{0pt}{0.2ex}}
\vskip-5ex
\begin{eqnarray*}
\text{\ajKakko{1}}\quad x+\myfrac{1}{\,x\,}
	&=& \myfrac{\,3-\sqrt{5}\,}{2}+\myfrac{\,3+\sqrt{5}\,}{2} \\
	&=& \myfrac{\,3-\sqrt{5}\, + \,3+\sqrt{5}}{2} \\
	&=& \myfrac{\,6\,}{2} \\
	&=& \, 3 \quad\text{\□答}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( x=\tfrac{\,3-\sqrt{5}\,}{2},\ 
	\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}
	=\tfrac{\,3+\sqrt{5}\,}{2},\ 
	x+\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}=3 \)\quad だよ
	\rule[-1ex]{0pt}{0.2ex}}
\vskip-5ex
\begin{eqnarray*}
\text{\ajKakko{2}}\quad x^2+\myfrac{1}{\,x^2}
	&=& \Bigl(x+\myfrac{1}{\,x\,}\Bigr)^2-2x\sshasen\cdot\myfrac{1}{\,x\sshasen\,} \\
	&=& \Bigl(x+\myfrac{1}{\,x\,}\Bigr)^2-2 \quad
		\text{\textcolor{CUDRed}{\tiny \ajLeftBArrow よく出てくる}}\\
	&=& \hooklen{\Bigl(x+\myfrac{1}{\,x\,}\Bigr)^2}{3^2}-2 \\
	&=& \hskip0.12\textwidth 9-2 \quad=7 \quad\text{\□答}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\setbeamercolor{frametitle}{bg=CUDCream!75,fg=black!80}
\setbeamercolor{palette primary}{bg=CUDCream!75,fg=black!80}

\begin{frame}[t]
\frametitle{\( x=\tfrac{\,3-\sqrt{5}\,}{2},
	\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}
	=\tfrac{\,3+\sqrt{5}\,}{2},
	x+\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}=3,
	x^2+\tfrac{\,1\,}{\,\raisebox{0.2ex}{\(\scriptstyle x^2\)}}=7 \)
	\rule[-1ex]{0pt}{0.2ex}}
\vskip-1ex
\ajKakko{3}\ \( x^2-\myfrac{1}{\,x^2} \)\ の事前準備で
\( x-\myfrac{1}{\,x\,} \)を計算します。

\vskip-2ex
\begin{eqnarray*}
& & \hspace{-2.4zw} x-\myfrac{1}{\,x\,} \\
&=& \myfrac{3-\sqrt{5}}{2}-\myfrac{3+\sqrt{5}}{2}
	= \myfrac{3-\sqrt{5}-(3+\sqrt{5})}{2}\\
	&=& \myfrac{3-\sqrt{5}-3-\sqrt{5}}{2}
	=\myfrac{-2\sqrt{5}}{2}=-\sqrt{5}\ \ittan
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\setbeamercolor{frametitle}{use=structure,fg=white,bg=structure.fg}
\setbeamercolor{palette primary}{use=structure,fg=white,bg=structure.fg}

\begin{frame}[t]
\frametitle{\( x+\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}=3,\quad
	x^2+\tfrac{\,1\,}{\,\raisebox{0.2ex}{\(\scriptstyle x^2\)}}=7,\quad
	x-\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}=-\sqrt{5} \)
	\rule[-1ex]{0pt}{0.2ex}}
\vskip-4ex
\begin{eqnarray*}
\text{\ajKakko{3}}\quad x^2-\myfrac{1}{\,x^2}
	&=& \Bigl(x+\myfrac{1}{\,x\,}\Bigr)
		\Bigl(x-\myfrac{1}{\,x\,}\Bigr)\\
	&=& \hooklen{\Bigl(x+\myfrac{1}{\,x\,})}{3}
		\cdot(-\sqrt{5})\\
	&=& \qquad -3\sqrt{5}\quad\text{\□答}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( x+\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}=3,\quad
	x^2+\tfrac{\,1\,}{\,\raisebox{0.2ex}{\(\scriptstyle x^2\)}}=7,\quad
	x-\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}=-\sqrt{5} \)
	\rule[-1ex]{0pt}{0.2ex}}
\vskip-5ex
\begin{eqnarray*}
\text{\ajKakko{4}}\quad x^4+\myfrac{1}{\,x^4}
	&=& (x^2)^2+\Bigl(\myfrac{1}{\,x^2}\Bigr)^2 \\
	&=& \Bigl(x^2+\myfrac{1}{\,x^2}\Bigr)^2
		-2x^2\ssshasen\cdot\myfrac{1}{\,x^2\ssshasen}\\
	&=& \Bigl(x^2+\myfrac{1}{\,x^2}\Bigr)^2	-2 \\
	&=& \hooklen{\Bigl(x^2+\myfrac{1}{\,x^2}\Bigr)^2}{7^2}-2
	\quad =47\quad\text{\□答}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( %x+\tfrac{\,1\,}{\raisebox{0.2ex}{\(\scriptstyle x\)}}=3,\ 
	x^2+\tfrac{\,1\,}{\,\raisebox{0.2ex}{\(\scriptstyle x^2\)}}=7,\ 
	x^2-\tfrac{\,1\,}{\,\raisebox{0.2ex}{\(\scriptstyle x^2\)}}=-3\sqrt{5} \)
	\rule[-1ex]{0pt}{0.2ex}}
\vskip-4ex
\begin{eqnarray*}
\text{\ajKakko{5}}\quad x^4-\myfrac{1}{\,x^4}
	&=& \Bigl(x^2+\myfrac{1}{\,x^2}\Bigr)
		\Bigl(x^2-\myfrac{1}{\,x^2}\Bigr)\\
	&=& \hooklen{\Bigl(x^2+\myfrac{1}{\,x^2})}{7}
		\cdot(-3\sqrt{5})\\
	&=& \qquad -21\sqrt{5}\quad\text{\□答}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\setbeamercolor{frametitle}{bg=CUDPink!75,fg=black!80}
\setbeamercolor{palette primary}{bg=CUDPink!75,fg=black!80}

\begin{frame}[t]
\frametitle{他にこんなのもよく出るよ}
\vskip-4.4ex
\begin{eqnarray*}
x^3+\myfrac{1}{\,x^3}
	&=& \Bigl(x+\myfrac{\,1\,}{x}\Bigr)^3-3x\sshasen\cdot
		\myfrac{1}{\,x\sshasen\,}\Bigl(x+\myfrac{\,1\,}{x}\Bigr) \\
	&=& \Bigl(x+\myfrac{\,1\,}{x}\Bigr)^3-3\Bigl(x+\myfrac{\,1\,}{x}\Bigr)
\end{eqnarray*}
\vskip-0.5ex
%%%%%%%\hrulefill
\hrule height0.8pt width \textwidth
\vskip-0.5ex
\begin{eqnarray*}
x^3-\myfrac{1}{\,x^3}
	&=& \Bigl(x-\myfrac{\,1\,}{x}\Bigr)^3+3x\sshasen\cdot
		\myfrac{1}{\,x\sshasen\,}\Bigl(x-\myfrac{\,1\,}{x}\Bigr) \\
	&=& \Bigl(x-\myfrac{\,1\,}{x}\Bigr)^3+3\Bigl(x-\myfrac{\,1\,}{x}\Bigr)
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{他にこんなのもよく出るよ}
\vskip-4ex
\begin{eqnarray*}
x^5+\myfrac{1}{\,x^5}
	&=& \Bigl(x^3+\myfrac{1}{\,x^3}\Bigr)\Bigl(x^2+\myfrac{1}{\,x^2}\Bigr)
		-\Bigl(x+\myfrac{1}{\,x}\Bigr) 
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\end{document}
