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方框方程，但只框出某些边

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方框方程，但只框出某些边

我需要框出方程的一部分，但我需要框的一侧是开放的。在下面的 MWE 中，我需要方程的左侧部分没有框的右侧框架，方程的右侧部分没有框的左侧框架。

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}
\begin{document}
This is my equation $\boxed{q_1^{\ast}}=\dfrac{26-2\cdot 1 -4\cdot 2}{2\cdot 1}=\boxed{8}$
\end{document}

答案1

以下 MWE 提供\lboxed和，\rboxed它们是截断的。从技术上讲，这与仅删除了右/左的部分\fbox相同（的定义取自\fbox\vrule\fboxlatex.ltx)。它还提供\llboxed和，\rrboxed它们是array样式框，其中所需的边剥离了垂直规则。

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\newcommand{\lboxed}[1]{\begin{array}{|l}\hline#1\\\hline\end{array}}
\newcommand{\rboxed}[1]{\begin{array}{r|}\hline#1\\\hline\end{array}}
\makeatletter
\newcommand{\llboxed}[1]{{%
  \let\@frameb@x\l@frameb@x\fbox{#1}%
}}
\newcommand{\rrboxed}[1]{{%
  \let\@frameb@x\r@frameb@x\fbox{#1}%
}}

\def\l@frameb@x#1{%
  \@tempdima\fboxrule
  \advance\@tempdima\fboxsep
  \advance\@tempdima\dp\@tempboxa
  \hbox{%
    \lower\@tempdima\hbox{%
      \vbox{%
        \hrule\@height\fboxrule
        \hbox{%
          \vrule\@width\fboxrule
          #1%
          \vbox{%
            \vskip\fboxsep
            \box\@tempboxa
            \vskip\fboxsep}%
          #1%
          %\vrule\@width\fboxrule% Right \vrule removed
                    }%
        \hrule\@height\fboxrule}%
                          }%
        }%
}
\def\r@frameb@x#1{%
  \@tempdima\fboxrule
  \advance\@tempdima\fboxsep
  \advance\@tempdima\dp\@tempboxa
  \hbox{%
    \lower\@tempdima\hbox{%
      \vbox{%
        \hrule\@height\fboxrule
        \hbox{%
          %\vrule\@width\fboxrule% Left \vrule removed
          #1%
          \vbox{%
            \vskip\fboxsep
            \box\@tempboxa
            \vskip\fboxsep}%
          #1%
          \vrule\@width\fboxrule}%
        \hrule\@height\fboxrule}%
                          }%
        }%
}
\makeatother
\begin{document}
This is my equation $\boxed{q_1^{\ast}}=\dfrac{26-2\cdot 1 -4\cdot 2}{2\cdot 1}=\boxed{8}$

This is my equation $\rboxed{q_1^{\ast}}=\dfrac{26-2\cdot 1 -4\cdot 2}{2\cdot 1}=\lboxed{8}$

This is my equation $\rrboxed{$q_1^{\ast}$}=\dfrac{26-2\cdot 1 -4\cdot 2}{2\cdot 1}=\llboxed{$8$}$
\end{document}

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