我如何从顶部消除这个偏移？

Question

\documentclass[14pt,a4paper,oneside]{extarticle}
\usepackage[left=30mm,right=15mm,top=20mm,bottom=20mm,bindingoffset=0mm]{geometry}
\setcounter{page}{2}
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\usepackage[english,russian]{babel}
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\usepackage{graphicx}
\usepackage{float}
\usepackage{amsmath}
\usepackage{colortbl}
%usepackage{setspace}
\usepackage{multicol}
\usepackage{indentfirst}
%usepackage{ccaption}
\usepackage{tocloft}
\usepackage{caption}
%captiondelim{. }

\setlength{\parindent}{1.25cm}
\renewcommand{\thesection}{\arabic{section}.}
\renewcommand{\thesubsection}{\arabic{section}.\arabic{subsection}.}
\renewcommand{\thesubsubsection}{\arabic{section}.\arabic{subsection}.\arabic{subsubsection}.}

\makeatletter
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                                     {4ex\@plus 1ex \@minus .2ex}%
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\renewcommand{\cftsecleader}{\cftdotfill{\cftdotsep}}

\DeclareCaptionFormat{CustomTable}{#2#1\\#3}
\DeclareCaptionLabelSeparator{fill}{\hfill}
\DeclareCaptionLabelFormat{fullparents}{\bothIfFirst{#1}{~}#2}
\captionsetup[table]{
     format=CustomTable,
     font={normalsize},
     labelformat=fullparents,
     labelsep=fill,
     justification=centering,
     singlelinecheck=false,
     skip=0pt
     }
     \theoremstyle{definition}
\newtheorem{definition}{Определение}
\newtheorem{theorem}{Теорема}
\newtheorem{proposition}{Следствие}
\newtheorem{lemma}{Лемма}

\sloppy
\begin{document}

\textbf{Решение}
\begin{enumerate} 
\item Гамильтониан: $H = \psi_1 v + \psi_2 (\frac{uF}{M} - g)$
\item Уравнения Эйлера-Лагранжа:
\begin{multicols}{3}
\[
        \begin{cases}
            \psi^{\prime}_1 = -\frac{\partial H}{\partial h}\\
            \psi^{\prime}_2 = -\frac{\partial H}{\partial v}\\
        \end{cases}
\]
\columnbreak
\[
        \begin{cases}
            \psi^{\prime}_1 = 0\\
            \psi^{\prime}_2 = -\psi_1\\
        \end{cases}
\]
\columnbreak    
\[
        \begin{cases}
            \psi_1 = C_1\\
            \psi_2 = -C_1 t + C_2\\
        \end{cases}
\]
\end{multicols}
\item  Условие принципа максимума Понтрягина:
$$maxH = max(\psi_1 v + \psi_2 (\frac{uF}{M} - g)) = \psi_1 v - \psi_2 g + max(\psi_2\frac{uF}{M})$$
\end{enumerate}

\end{document}

我如何从顶部消除这个偏移？

Answer 1

您的最小工作示例不完整。尽管如此，我还是设法重现了错误。您必须\noindent在第一列的开头添加一个。

在我看来，通过 multicol 的方式相当不方便。因此，我把所有这些都简单地放到了 parboxes 中

\begin{enumerate} 

\item DEF: $H = \psi_1 v + \psi_2 (\frac{uF}{M} - g)$ 
\item 123 456:\\
\parbox[c]{0.4\linewidth}{
\[
\begin{cases}
\psi^{\prime}_1 = - \frac{\partial H}{\partial h}\\ \psi^{\prime}_2 = - \frac{\partial H}{\partial v} \end{cases}
\] 
}\parbox[c]{0.3\linewidth}{
\[ 
\begin{cases} \psi^{\prime}_1 = 0\\ \psi^{\prime}_2 = -\psi_1 
\end{cases} \] 
}\parbox[c]{0.3\linewidth}{
\[ 
\begin{cases} \psi_1 = C_1\\ \psi_2 = -C_1 t + C_2 
\end{cases} \]
}%parbox
\item GHI JKL MNO PQR: $$maxH = max(\psi_1 v + \psi_2 (\frac{uF}{M} - g)) = \psi_1 v - \psi_2 g + max(\psi_2\frac{uF}{M})$$ 
\end{enumerate}

Answer 2

您不需要multicols环境parbox。只需在一行中写入方程式，并\qquad在其间留出间隙cases：

\documentclass[12pt, a4paper]{extarticle}
\usepackage{mathtools}
\usepackage{enumitem}

\begin{document}
\begin{enumerate}

\item DEF: $H = \psi_1 v + \psi_2 (\frac{uF}{M} - g)$
\item 123 456:\\
\[
\begin{cases}
\psi^{\prime}_1 = - \frac{\partial H}{\partial h}\\ 
\psi^{\prime}_2 = - \frac{\partial H}{\partial v} 
\end{cases}
\qquad
\begin{cases} 
\psi^{\prime}_1 = 0\\ 
\psi^{\prime}_2 = -\psi_1
\end{cases} 
\qquad
\begin{cases} 
\psi_1 = C_1\\ 
\psi_2 = -C_1 t + C_2
\end{cases} 
\]
\item GHI JKL MNO PQR: 
\[
\max{H} = \max(\psi_1 v + \psi_2 \biggl(\frac{uF}{M} - g)\biggr)
        = \psi_1 v - \psi_2 g + \max\biggl(\psi_2\frac{uF}{M}\biggr)
\]
\end{enumerate}
\end{document}

在上面的 MWE 中，我纠正了最后一个等式，因为我认为它应该正确写出。