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按照左右箭头符号排列方程式

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按照左右箭头符号排列方程式

式中的符号后面怎样排列Leftrightarrow?

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{float}
\usepackage{graphics}

\begin{document}
\begin{equation*}
    \begin{aligned}
        & \Leftrightarrow - \scalebox{0.975}{$\biggl(\left(\alpha + \mu + \rho\right)\left(\left( a_1 + \left( \alpha + \mu\right)\right)^2 + a_1 a_2\right) + \left( a_1 + \left(\alpha + \mu\right)\right)\gamma \rho \biggr)$} > 0 \\
        & \Leftrightarrow \left(\alpha + \mu + \rho\right)\left(\left( a_1 + \left( \alpha + \mu\right)\right)^2 + a_1 a_2\right) + \left( a_1 + \left(\alpha + \mu\right)\right)\gamma \rho < 0 \\
        & \Leftrightarrow \scalebox{0.875}{$\left(\alpha + \mu + \rho\right)\left(\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2 + \Lambda \beta - \frac{\left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right)}{\alpha+\mu+\rho}\right) + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho$} < 0 \\
        & \Leftrightarrow \scalebox{0.85}{$\left(\alpha + \mu + \rho\right)\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2+ \Lambda \beta\left(\alpha + \mu + \rho\right) - \left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right) $} \\
        & \scalebox{0.9}{$ + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho $} < 0 \\
        & \Leftrightarrow \scalebox{0.9}{$\left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right)$} > \\
        & \scalebox{0.9}{$\left(\alpha + \mu + \rho\right)\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2+ \Lambda \beta\left(\alpha + \mu + \rho\right) + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho$} \\
    \end{aligned}
\end{equation*}
\end{document}

答案1

您可以使用包IEEEeqnarray中的环境IEEEtrantools来实现良好的对齐。

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{IEEEtrantools}
\usepackage{float}
\usepackage{graphics}
\usepackage{showframe}

\begin{document}
Using \verb|\IEEEeqnarray| with \verb|\scalebox|
\begin{IEEEeqnarray*}{rCl}
        & \Leftrightarrow & - \scalebox{0.975}{$\biggl(\left(\alpha + \mu + \rho\right)\left(\left( a_1 + \left( \alpha + \mu\right)\right)^2 + a_1 a_2\right) + \left( a_1 + \left(\alpha + \mu\right)\right)\gamma \rho \biggr)$} > 0 \\
        & \Leftrightarrow & \left(\alpha + \mu + \rho\right)\left(\left( a_1 + \left( \alpha + \mu\right)\right)^2 + a_1 a_2\right) + \left( a_1 + \left(\alpha + \mu\right)\right)\gamma \rho < 0 \\
        & \Leftrightarrow & \scalebox{0.875}{$\left(\alpha + \mu + \rho\right)\left(\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2 + \Lambda \beta - \frac{\left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right)}{\alpha+\mu+\rho}\right) + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho$} < 0 \\
        & \Leftrightarrow & \scalebox{0.85}{$\left(\alpha + \mu + \rho\right)\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2+ \Lambda \beta\left(\alpha + \mu + \rho\right) - \left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right) $} \\
        & & \scalebox{0.9}{$ + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho $} < 0 \\
        & \Leftrightarrow & \scalebox{0.9}{$\left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right)$} > \\
        & & \scalebox{0.9}{$\left(\alpha + \mu + \rho\right)\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2+ \Lambda \beta\left(\alpha + \mu + \rho\right) + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho$} \\
\end{IEEEeqnarray*}

Using \verb|\IEEEeqnarray| with \verb|\footnotesize|
{\footnotesize
\begin{IEEEeqnarray*}{rCl}
        & \Leftrightarrow & - \biggl(\left(\alpha + \mu + \rho\right)\left(\left( a_1 + \left( \alpha + \mu\right)\right)^2 + a_1 a_2\right) + \left( a_1 + \left(\alpha + \mu\right)\right)\gamma \rho \biggr) > 0 \\
        & \Leftrightarrow & \left(\alpha + \mu + \rho\right)\left(\left( a_1 + \left( \alpha + \mu\right)\right)^2 + a_1 a_2\right) + \left( a_1 + \left(\alpha + \mu\right)\right)\gamma \rho < 0 \\
        & \Leftrightarrow & \left(\alpha + \mu + \rho\right)\left(\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2 + \Lambda \beta - \frac{\left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right)}{\alpha+\mu+\rho}\right) \\
        & & \negmedspace {} + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho < 0 \\
        & \Leftrightarrow & \left(\alpha + \mu + \rho\right)\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2+ \Lambda \beta\left(\alpha + \mu + \rho\right) - \left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right)  \\
        & & \negmedspace {} + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho  < 0 \\
        & \Leftrightarrow & \left(\alpha+ \mu\right)^2\left(\alpha + \mu + \gamma + \rho\right) > \left(\alpha + \mu + \rho\right)\left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)^2\\
        & & \negmedspace {} + \Lambda \beta\left(\alpha + \mu + \rho\right) + \left(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\right)\gamma \rho \\
\end{IEEEeqnarray*}}
\end{document}

在此处输入图片描述

注意：您可以阅读使用 \left 和 \right 是否有坏处？。此外，不鼓励包含文本的缩放元素。

答案2

我强烈建议你不要使用\adustbox 任何地方在此数学表达式中。如果需要，只需手动插入几个换行符即可。

我还会使用align*环境而不是嵌套equation*/aligned设置。

单独的观察：这对你来说可能是一次宣泄的经历不是在每个括号上使用\leftand \right。我可以数出这两个自动调整大小指令各有 50 多个 [!] 实例。通过一些相当简单的分析，这个数字可以减少到 9 [是的，只有 9] 对“大”括号。并且，不要只使用圆括号，也不要时不时地考虑使用方括号（或花括号......）来提供一些简单但非常有效的视觉线索。

在此处输入图片描述

\documentclass{article}
%\usepackage[utf8]{inputenc} % that's the default nowadays
\usepackage[T1]{fontenc}
\usepackage{amsmath,amssymb}

\begin{document}
\begin{align*}
  &\Leftrightarrow -\bigl[
    (\alpha + \mu + \rho)\bigl(( a_1 + ( \alpha + \mu))^2 + a_1 a_2\bigr) 
    + \bigl( a_1 + (\alpha + \mu)\bigr)\gamma \rho \bigr] > 0 \\
  &\Leftrightarrow 
   (\alpha + \mu + \rho)\bigl(( a_1 + ( \alpha + \mu))^2 + a_1 a_2\bigr) 
    + \bigl( a_1 + (\alpha + \mu)\bigr)\gamma \rho < 0 \\
  &\Leftrightarrow 
   (\alpha + \mu + \rho)\biggl[\biggl(\frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}\biggr)^{\!\!2} + \Lambda \beta 
  - \frac{(\alpha+ \mu)^2(\alpha + \mu + \gamma + \rho)}{\alpha+\mu+\rho}\biggr] \\
  &\qquad + \frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}\,\gamma \rho < 0 \\
  &\Leftrightarrow (\alpha + \mu + \rho)
   \biggl(\frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}\biggr)^{\!\!2}
   + \Lambda \beta(\alpha + \mu + \rho) \\
  &\qquad - (\alpha+ \mu)^2(\alpha + \mu + \gamma + \rho)  
   + \frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}\,\gamma \rho  < 0 \\
  &\Leftrightarrow (\alpha+ \mu)^2(\alpha + \mu + \gamma + \rho) >{} \\
  &\qquad (\alpha + \mu + \rho)\biggl(\frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}\biggr)^{\!\!2} 
  + \Lambda \beta(\alpha + \mu + \rho) \\
  &\qquad\quad
  + \frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}\,\gamma \rho 
\end{align*}
\end{document}

答案3

使用包multlined中定义的mathtools，无需缩放方程系统的部分，也无需手动设置括号的大小：

\documentclass{article}
\usepackage{mathtools}

%--------------- show page layout. don't use in a real document!
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%

\begin{document}
\[
    \begin{aligned}
\Leftrightarrow 
    & - \biggl((\alpha + \mu + \rho) \bigl((a_1 + \alpha + \mu)^2 + a_1 a_2\bigr) + ( a_1 + \alpha + \mu)\gamma \rho \biggr) > 0 \\
\Leftrightarrow {}
    & (\alpha + \mu + \rho)\bigl(( a_1 + \alpha + \mu)^2 + a_1 a_2\bigr) + (a_1 + \alpha + \mu)\gamma \rho < 0 \\
\Leftrightarrow {}
    & \begin{multlined}[t]
        (\alpha + \mu + \rho)
        \Biggl(\biggl(\frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}
               \biggr)^2    \\
        + \Lambda \beta - \frac{(\alpha+ \mu)^2 (\alpha + \mu + \gamma + \rho)}{\alpha+\mu+\rho}\Biggr)
         + \biggl(\frac{\Lambda\beta \left(\alpha+\mu+\rho\right)}{\left(\alpha+ \mu\right)\left(\alpha + \mu + \gamma + \rho\right)}\biggr)\gamma \rho  < 0
     \end{multlined}\\
\Leftrightarrow {}
    & \begin{multlined}[t]
        (\alpha + \mu + \rho)
        \biggl(\frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}\biggr)^2
        + \Lambda \beta (\alpha + \mu + \rho )     \\
        -  (\alpha+ \mu)^2 (\alpha + \mu + \gamma + \rho)
        + \biggl(\frac{\Lambda\beta (\alpha+\mu+\rho )}{(\alpha+ \mu) (\alpha + \mu + \gamma + \rho)}\biggr)\gamma \rho < 0  
     \end{multlined}    \\
\Leftrightarrow {}
    & \begin{multlined}[t]
        (\alpha+ \mu)^2 (\alpha + \mu + \gamma + \rho) >    \\
        (\alpha + \mu + \rho)
        \biggl(\frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)
        (\alpha + \mu + \gamma + \rho)}\biggr)^2            \\
        +  \Lambda \beta(\alpha + \mu + \rho) + 
        \biggl(\frac{\Lambda\beta (\alpha+\mu+\rho)}{(\alpha+ \mu)(\alpha + \mu + \gamma + \rho)}\biggr)\gamma \rho
     \end{multlined}
    \end{aligned}
\]
\end{document}

在此处输入图片描述

（红线表示文本边框）

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