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如何修复多行方程中过满的 \hbox

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如何修复多行方程中过满的 \hbox 
\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\addtolength{\oddsidemargin}{-.5in}
\addtolength{\evensidemargin}{1in}
\addtolength{\textwidth}{1.15in}
\addtolength{\topmargin}{-.75in}
\addtolength{\textheight}{1in}
\begin{document}
\begin{align}
    \nonumber &E_{AH}^* = (S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) = \Big(\frac{\mu_b + \delta_b}{\beta_B}, \frac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)},\\ \nonumber &\frac{\Lambda_h\beta_B(\mu_b+\delta_b)}{\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack},
    \frac{\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack},\\
     &\frac{\gamma_a\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{\mu_h(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack}\Big). \label{Eqn: EEAH}
\end{align}
\end{document}

输出如下所示。 第二行已经满了

有没有办法来解决这个问题？

答案1

公式的长度超出了允许的水平空间，为了便于阅读，将其分成五行。拆分行时，应具有与符号=在所有行中都存在时相同的虚拟空间。为此，请\phantom{{}={}}在没有 的行的开头插入=。此外，添加两个括号{}={}是因为=是一个二进制运算符，就像+或 一样-，这些括号充当数学原子，用于在符号周围保持正确的间距=。

\documentclass[letterpaper,12pt]{article}
\addtolength{\oddsidemargin}{-.5in}
\addtolength{\evensidemargin}{1in}
\addtolength{\textwidth}{1.15in}
\addtolength{\topmargin}{-.75in}
\addtolength{\textheight}{1in}
\usepackage{amsmath}
\begin{document}
\begin{align}
    \nonumber E_{AH}^* &= (S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) \\ \nonumber
    &= \Big(\frac{\mu_b + \delta_b}{\beta_B}, \frac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)},\\ \nonumber 
    &\phantom{{}={}} \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}{\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}, \\ \nonumber
    &\phantom{{}={}} \frac{\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack},\\
    &\phantom{{}={}} \frac{\gamma_a\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{\mu_h(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack}\Big). \label{Eqn: EEAH}
\end{align}
\end{document}

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答案2

你的读者会很难找到符号和它们的展开之间的对应关系。我建议把主要方程放在中心，然后解释它下面的符号。

\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}

\begin{document}

\begin{gather}
E_{AH}^* = (S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) \label{Eqn: EEAH} \\
\begin{align*}
S_b^* &= \frac{\mu_b + \delta_b}{\beta_B} \\
I_b^* &= \frac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)} \\
S_h^* &=
  \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}
       {\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}[\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)]}
  \\
I_a^* &=
  \frac{\Lambda_h\beta_{BH}[\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)]}
       {(\mu_h+d+\gamma_a)\bigl[\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}
        \bigl(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\bigr)\bigr]}
  \\
R_h^* &=
  \frac{\gamma_a\Lambda_h\beta_{BH}[\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)]}
       {\mu_h(\mu_h+d+\gamma_a)\bigl[\mu_h\beta_B(\mu_b+\delta_b)+
        \beta_{BH}\bigl(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\bigr)\bigr]}. 
\end{align*}
\end{gather}

\end{document}

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答案3

我将 (a) 引入一个换行符，并 (b) 将所有行相对于第一个=符号垂直对齐（有偏移）。我还将稍微增加连续行之间的空间，并将最外层括号的大小从\Big增加到\bigg。

在此处输入图片描述

\documentclass[letterpaper,12pt]{article}
\addtolength{\oddsidemargin}{-.5in}
\addtolength{\evensidemargin}{1in}
\addtolength{\textwidth}{1.15in}
\addtolength{\topmargin}{-.75in}
\addtolength{\textheight}{1in}
\usepackage{amsmath}
\begin{document}
\begin{align}
E_{AH}^* &= (S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) \notag \\[1ex]
&= \biggl(\frac{\mu_b + \delta_b}{\beta_B}, 
   \frac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)}, \notag \\[1.5ex]
&\qquad \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}{\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}, \notag \\[1.5ex]
&\qquad \frac{\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\bigl(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\bigr)\big\rbrack}, \notag \\[1.5ex]
&\qquad\frac{\gamma_a\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{\mu_h(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\bigl(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\bigr)\big\rbrack}\biggr). \label{Eqn: EEAH}
\end{align}
\end{document}

答案4

另外 4 行还有 3 种变体：

    \documentclass{article}
\usepackage[showframe]{geometry}
    \usepackage{mathtools,nccmath}
     \newcommand*{\dd}{\mathop{}\!\mathrm{d}}

    \begin{document}

\begin{equation}
\begin{alignedat}[b]{2}
    E_{AH}^*&=\mathrlap{(S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) } \\
     & = & \biggl(\frac{\mu_b + \delta_b}{\beta_B}, \mfrac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)}, \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}{\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack} & , \\
    & & \frac{\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack} & , \\
     & & \frac{\gamma_a\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{\mu_h(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack} &\biggr). \label{Eqn: EEAH}
\end{alignedat}
\end{equation}
\bigskip
\begin{equation}
\begin{alignedat}[b]{2}
    E_{AH}^*&=\mathrlap{(S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) } \\
     & = \begin{multlined}[t] \biggl(\frac{\mu_b + \delta_b}{\beta_B}, \mfrac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)}, \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}{\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack} , \\
    \frac{\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack} , \\
    \frac{\gamma_a\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{\mu_h(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack} \biggr).
     \end{multlined} \label{Eqn: EEAH}
\end{alignedat}
\end{equation}
\bigskip
\begin{fleqn}[2em]
\begin{equation}
\begin{alignedat}[b]{2}
    E_{AH}^*&=\mathrlap{(S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) } \\
     & = & \biggl(\frac{\mu_b + \delta_b}{\beta_B},& \frac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)}, \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}{\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack} , \\
    & & & \frac{\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack} , \\
     & & & \frac{\gamma_a\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{\mu_h(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack} \biggr). \label{Eqn: EEAH}
\end{alignedat}
\end{equation}
\end{fleqn}

    \end{document}

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