%tDB様

%毎度素早い回答ありがとうございます。

%今回のリストでは
%\usepackage[maskAnsfalse]{emathAe}
%としても，同じように右側に出力されました。

%しかし，はじめの大問に\kaitou{}を追加すると
%うまくいきました。
%使い方は正しいでしょうか？

%\usepackage[maskAnsfalse]{emathAe}を
%うまく使っていきます。
%ありがとうございました。
%-----------------------
\documentclass[b4j,fleqn]{jarticle}     
\usepackage[margin=8mm]{geometry}
\usepackage[papersize]{emathP}
\usepackage[maskAnsfalse]{emathAe}
\setlength{\columnseprule}{0.4pt}
\setlength{\columnsep}{2zw}
\pagestyle{empty}
\preEqlabel{$\cdots\cdots$}%
\resetcounter{equation}[enumi]
\def\testname{2005年度前期中間考査}
\def\testbi{2005.06.13}

\begin{document}
\enumSep{\narrowenumsep}
\enumLmargin{0.1zw}
\twocolumn[%
 \begin{flushright}
 \texttt{\testbi}
 \end{flushright}
 \begin{center}
\large\textgt{%
\testname%
\quad 1年・数学2}
 \end{center}
]%
\def\KaitouTTL{\footnotesize\medskip\par\empty}
\begin{Enumerate}[\protect\Large\expandafter\fbox 1]
\item 次の数を$i$を用いて表せ。\kaitou{} %%%%%%%%%<=ここに\kaitou{}を追加
\begin{edaenumerate}
<retusuu=3,gyoukan=3.5\baselineskip,edasikiri,%
edatopsep=-.8zh,%縦間隔
>[(1)]
\item $\sqrt{-9}$
\kaitou{$\bm{3i}$\kotae}
\item $\sqrt{-28}$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}%\intertext{}
=&{}\sqrt{28}i\\
=&{}\bm{2\sqrt{7}i}\kotae
\end{align*}}}
\item $\sqrt{-\bunsuu{25}{3}}$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}%\intertext{}
=&{}\sqrt{\bunsuu{25}{3}}i=\bunsuu{5}{\sqrt3}i\\
=&{}\bm{\bunsuu{5\sqrt3}{3}i}\kotae
\end{align*}}}
\end{edaenumerate}
\item 次の式を1つの複素数で表せ。
\begin{edaenumerate}
<gyoukan=8\baselineskip,edasikiri,edatopsep=-.5zh,%縦間隔
>[(1)]
\item $\sqrt{-4}-\sqrt{-9}$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}%\intertext{}
=&{}\sqrt{4}i-\sqrt{9}i=2i-3i\\
=&{}\bm{-i}\kotae
\end{align*}}}
\item $\sqrt{-2}\times \sqrt{-3}$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
=&{}\sqrt{2}i\times \sqrt{3}i=\sqrt6 i^2\\
=&{}\sqrt6\times (-1)=\bm{-\sqrt6}\kotae
\end{align*}}}
\item $\bunsuu{\sqrt{6}}{\sqrt{-3}}$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
=&{}\bunsuu{\sqrt{6}}{\sqrt{3}i}=%
\bunsuu{\sqrt{6}\times \sqrt{3}i}{\sqrt{3}i\times \sqrt{3}i}%
=\bunsuu{\sqrt{18}i}{3i^2}\\
=&{}\bunsuu{3\sqrt{2}i}{3\times (-1)}=\bm{-\sqrt2i}\kotae
\end{align*}}}
\item $\bunsuu{\sqrt{3}+\sqrt{-2}}{\sqrt{3}-\sqrt{-2}}$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
=&{}\bunsuu{\sqrt{3}+\sqrt{2}i}{\sqrt{3}-\sqrt{2}i}=%
\bunsuu{(\sqrt{3}+\sqrt{2}i)(\sqrt{3}+\sqrt{2}i)}{(\sqrt{3}-\sqrt{2}i)(\sqrt{3}\bm{+}\sqrt{2}i)}\\%
=&{}\bunsuu{3+2\sqrt6i+2i^2}{3-2i^2}\\
=&{}\bunsuu{3+2\sqrt6i+2\times (-1)}{3-2\times (-1)}\\
=&{}\bm{\bunsuu{1+2\sqrt6i}{5}}\kotae
\end{align*}}}
\item $(1+i)^2$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
=&{}1+2i+i^2=1+2i+(-1)\\
=&{}\bm{2i}\kotae
\end{align*}}}
\item $(\sqrt{2}+3i)(\sqrt{2}-3i)$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
=&{}(\sqrt{2})^2-(3i)^2=2-9i^2\\
=&{}2-9\times (-1)=\bm{11}\kotae
\end{align*}}}
\item $\bunsuu{1-i}{2+3i}$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
=&{}\bunsuu{(1-i)(2\bm{-}3i)}{(2+3i)(2\bm{-}3i)}%
=\bunsuu{2-3i-2i+3i^2}{2^2-(3i)^2}\\
=&{}\bunsuu{-1-5i}{13}=\bm{\bunsuu{-1}{6}-\bunsuu{5}{6}i}\kotae
\end{align*}}}
\item $1+\bunsuu1i+\bunsuu{1}{i^2}+\bunsuu{1}{i^3}$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
=&{}1+\bunsuu1i+\bunsuu{1}{-1}+\bunsuu{1}{-i}%
=1+\bunsuu1i-1-\bunsuu{1}{i}\\
=&{}\bm{0}\kotae
\end{align*}}}
\end{edaenumerate}
\item 次の方程式を因数分解を利用して解け。
\begin{edaenumerate}
<gyoukan=10\baselineskip,edasikiri,
edatopsep=-.5zh,%縦間隔
>[(1)]
\item $x^2+8x+12=0$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
&{}(x+2)(x+6)=0\\%
\bm{x=}&{}\bm{-2,-6}\kotae
\end{align*}}}
\item $3x^2-x-2=0$
\kaitou{%
{\mathindent=0zw
\mbox{}
\vspace*{-\baselineskip}%
\vspace*{-\abovedisplayskip}%
\begin{align*}
&\tasuki{3}{1}{2}{-1}\\
&{}(x+2)(x+6)=0\\%
\bm{x=}&{}\bm{-2,-6}\kotae
\end{align*}}}
\end{edaenumerate}
\bigskip
\newpage
\end{Enumerate}
\end{document}

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