两列中的方程比 tcolorbox 更宽

两列中的方程比 tcolorbox 更宽

我有一个很大的方程式,我想把它放在tcolorbox一页里\twocolumn而不分割它。你能帮我解决这个问题吗?

我甚至尝试了 \begin{minipage}[c]{5in} 但没有成功

平均能量损失

 \documentclass[a4paper,twocolumn]{article}


\usepackage{amsmath}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage[svgnames]{xcolor}
\usepackage[most]{tcolorbox}

\tcbset{
    lemmastyle/.style={enhanced, colback=white, colframe=blue!20, arc=0pt, 
                       fonttitle=\bfseries, description color=Maroon,  
                       colbacktitle=white, coltitle=DarkOliveGreen,    
                       top=\tcboxedtitleheight,
                       boxed title style={arc=0pt},
                       attach boxed title to top right={yshift=-\tcboxedtitleheight/3, 
                                                                                                            xshift=-2mm}%
                      },
}

\newtcbtheorem{myLemma}{Exercise}{lemmastyle}{thm}

\usepackage{pifont}


\begin{document}

\begin{myLemma}{}{}
\begin{enumerate}
\item Show that 
${\displaystyle D_2f(x,y) = \frac{\partial {}}{\partial{y}} \left ( \int_0^xg_1 (t,0) \ dt + \int_0^y g_2(x,s) \ ds \right)}$
\item prove that 
${ \displaystyle \left(\forall x\in\mathbb{R}  \right)\left(\forall y \in \mathbb{R} \right) x\neq y\, \text{and} \, x+y \neq 2 \implies x^{2}-2x \neq y^2-2y }$ 
\end{enumerate}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}


\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}


\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}


\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}

\begin{myLemma}{}{}
\end{myLemma}


\end{document}

在此处输入图片描述

答案1

只需删除外部括号对。它们在这里有什么用?如果你真的需要它们,可以分成几组。减少从框架到内部文本左边缘的距离,并将左边距上的枚举与包对齐enumitem。最后,如果你减小字体大小,请这样做外部数学模式:

    \documentclass[a4paper,twocolumn]{article}

    \usepackage{amsmath,amsfonts,amssymb}
    \usepackage[svgnames]{xcolor}
    \usepackage[most]{tcolorbox}

    \tcbset{
        lemmastyle/.style={enhanced, colback=white, colframe=blue!20, arc=0pt,
                           fonttitle=\bfseries, description color=Maroon,
                           colbacktitle=white, coltitle=DarkOliveGreen,
                           top=\tcboxedtitleheight,
                           boxed title style={arc=0pt},
                           attach boxed title to top right={yshift=-\tcboxedtitleheight/3, xshift=-2mm}%
                          }, left=0mm
    }

    \newtcbtheorem{myLemma}{Exercise}{lemmastyle}{thm}

    \usepackage{enumitem}

    \begin{document}

    \begin{myLemma}{}{}
\begin{enumerate}[wide=0pt, leftmargin=*]
\item Show that\\\small
${\tiny \displaystyle D_2f(x,y) = \frac{\partial {}}{\partial{y}} \biggl( \int_0^xg_1 (t,0) \ dt + \int_0^y g_2(x,s) \ ds \biggr)}$
\item Prove that\\\small
$\displaystyle \left(\forall x\in\mathbb{R} \right)\left(\forall y \in \mathbb{R} \right) x\neq y\; \text{and} \; x+y \neq 2 \implies x^{2}-2x \neq y^2-2y $
\end{enumerate}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}


    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}


    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}


    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \begin{myLemma}{}{}
    \end{myLemma}

    \end{document} 

在此处输入图片描述

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