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对齐一组四个方程（在左侧）

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对齐一组四个方程（在左侧）

问题：我有一组四个方程（其中一些很长）我该如何对齐所有这些方程。

平均能量损失

\documentclass[12pt,a4paper,fleqn]{article}
\usepackage[T1]{fontenc}
\usepackage[margin=1cm]{geometry}
\usepackage{amssymb,mathtools,amsthm}
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{parskip}
\usepackage{xcolor}
\usepackage{cellspace}
\setlength\cellspacetoplimit{4pt}
\setlength\cellspacebottomlimit{4pt}
\usepackage{nomencl}
\usepackage{setspace}
\setstretch{1.5}
\begin{document}
\begin{equation}
    \begin{split}
        (1+K)F''' +F'' F - \left(\dfrac{2n}{n+1}\right)\,F'\,^{2} -\tau \left(F'+\frac{1}{2}\,\eta\,F''\right)+\dfrac{2\lambda}{n+1}\,\left(\theta+\widetilde{N}C\right) \\[2ex]
        -\dfrac{2}{n+1}\left(M^{2}+\dfrac{1}{k_{1}}\right)F'+ KG'=0
    \end{split}
\end{equation}
\begin{flalign}
    \left(1+\dfrac{K}{2}\right)G'' +FG'-\tau \left(\frac{3}{2}G+\frac{1}{2}\,\eta\,G\right)-\left(\dfrac{3n-1}{n+1}\right)F'G-\left(\dfrac{2K}{n+1}\right)\left(2G+F''\right)&=0&
\end{flalign}
\begin{flalign}
    \begin{split}
        \left(1+\dfrac{4}{3}\,N_{r}\right)\theta''\,-P_{r}\,\tau\left(2\theta+\frac{1}{2}\,\eta\theta\right)\,+\,P_{r}\left(F\,\theta'-F'\,\theta\right)\,+\,P_{r}\cdot E_{c}\,\left(1+K\right)\,F''\,^{2}\\[2ex]
        +P_{r}\cdot E_{c}\cdot M^{2}F'\,^{2}+A^{\ast}\,F'+B^{\ast}\theta=0
    \end{split}
\end{flalign}
\begin{flalign}
    C'' +S_{c}\left(F\,C'-C\,F'\right)-S_{c}\left(C+\frac{1}{2}\,\eta\,C'\right)+S_{c}\cdot S_{r}\,\theta''-K_{r}\cdot S_{c}&=0&
\end{flalign}
\end{document}

答案1

我认为您并不真正需要几个连续的环境。使用 时要保持一致\,：您的大多数用法实际上都不正确。请注意F'^{\,2}不要使用可疑的F'\,^{2}。

\documentclass[12pt,a4paper,fleqn]{article}
\usepackage[T1]{fontenc}
\usepackage[margin=1cm]{geometry}
\usepackage{amssymb,mathtools,amsthm}
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{parskip}
\usepackage{xcolor}
\usepackage{cellspace}
\setlength\cellspacetoplimit{4pt}
\setlength\cellspacebottomlimit{4pt}
\usepackage{nomencl}
\usepackage{setspace}
\setstretch{1.5}

\begin{document}

With “left-right” alignment
\begin{align}
&\begin{aligned}
  (1+K)F''' +
  F'' F - 
  \left(\frac{2n}{n+1}\right)F'^{\,2} -
  \tau \left(F'+\frac{1}{2}\eta F''\right) +
  \dfrac{2\lambda}{n+1}\left(\theta+\widetilde{N}C\right)
  \\
  {}-\dfrac{2}{n+1}\left(M^{2}+\dfrac{1}{k_{1}}\right)F'+ KG'=0
\end{aligned}
\\[2ex]
& \left(1+\dfrac{K}{2}\right)G'' +
  FG' -
  \tau \left(\frac{3}{2}G +
  \frac{1}{2}\eta G\right) -
  \left(\dfrac{3n-1}{n+1}\right)F'G -
  \left(\dfrac{2K}{n+1}\right)(2G+F'')=0
\\[2ex]
&\begin{aligned}
 \left(1+\dfrac{4}{3}N_{r}\right)\theta'' -
 P_{r}\tau\left(2\theta+\frac{1}{2}\eta\theta\right) +
 P_{r}\left(F\theta'-F'\theta\right) +
 P_{r}\cdot E_{c}\left(1+K\right)F''^{\,2}
 \\
 {}+P_{r}\cdot E_{c}\cdot M^{2}F'^{\,2} +
 A^{\ast}F' +
 B^{\ast}\theta=0
 \end{aligned}
\\[2ex]
& C'' +
  S_{c}\left(FC'-CF'\right) -
  S_{c}\left(C+\frac{1}{2}\eta C'\right) +
  S_{c}\cdot S_{r}\theta'' -
  K_{r}\cdot S_{c}=0
\end{align}

A more conventional choice
\begin{align}
&\begin{multlined}
  (1+K)F''' +
  F'' F - 
  \left(\frac{2n}{n+1}\right)F'^{\,2} -
  \tau \left(F'+\frac{1}{2}\eta F''\right) +
  \dfrac{2\lambda}{n+1}\left(\theta+\widetilde{N}C\right)
  \\
  -\dfrac{2}{n+1}\left(M^{2}+\dfrac{1}{k_{1}}\right)F'+ KG'=0
\end{multlined}
\\[2ex]
& \left(1+\dfrac{K}{2}\right)G'' +
  FG' -
  \tau \left(\frac{3}{2}G +
  \frac{1}{2}\eta G\right) -
  \left(\dfrac{3n-1}{n+1}\right)F'G -
  \left(\dfrac{2K}{n+1}\right)(2G+F'')=0
\\[2ex]
&\begin{multlined}
 \left(1+\dfrac{4}{3}N_{r}\right)\theta'' -
 P_{r}\tau\left(2\theta+\frac{1}{2}\eta\theta\right) +
 P_{r}\left(F\theta'-F'\theta\right) +
 P_{r}\cdot E_{c}\left(1+K\right)F''^{\,2}
 \\
 +P_{r}\cdot E_{c}\cdot M^{2}F'^{\,2} +
 A^{\ast}F' +
 B^{\ast}\theta=0
 \end{multlined}
\\[2ex]
& C'' +
  S_{c}\left(FC'-CF'\right) -
  S_{c}\left(C+\frac{1}{2}\eta C'\right) +
  S_{c}\cdot S_{r}\theta'' -
  K_{r}\cdot S_{c}=0
\end{align}

\end{document}

在此处输入图片描述

答案2

这就是你要找的人吗？

\documentclass[12pt,a4paper,fleqn]{article}
\usepackage[T1]{fontenc}
\usepackage[margin=1cm]{geometry}
\usepackage{amssymb,mathtools,amsthm}
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{parskip}
\usepackage{xcolor}
\usepackage{cellspace}
\setlength\cellspacetoplimit{4pt}
\setlength\cellspacebottomlimit{4pt}
\usepackage{nomencl}
\usepackage{setspace}
\setstretch{1.5}
\begin{document}
\begin{align}
    \begin{split}
&        (1+K)F''' +F'' F - \left(\dfrac{2n}{n+1}\right)\,F'\,^{2} -\tau \left(F'+\frac{1}{2}\,\eta\,F''\right)+\dfrac{2\lambda}{n+1}\,\left(\theta+\widetilde{N}C\right) \\[2ex]
&\quad        -\dfrac{2}{n+1}\left(M^{2}+\dfrac{1}{k_{1}}\right)F'+ KG'=0
    \end{split}
\\
\begin{split}
    \left(1+\dfrac{K}{2}\right)G'' +FG'-\tau \left(\frac{3}{2}G+\frac{1}{2}\,\eta\,G\right)-\left(\dfrac{3n-1}{n+1}\right)F'G-\left(\dfrac{2K}{n+1}\right)\left(2G+F''\right)=0
\end{split}
\\
    \begin{split}
&        \left(1+\dfrac{4}{3}\,N_{r}\right)\theta''\,-P_{r}\,\tau\left(2\theta+\frac{1}{2}\,\eta\theta\right)\,+\,P_{r}\left(F\,\theta'-F'\,\theta\right)\,+\,P_{r}\cdot E_{c}\,\left(1+K\right)\,F''\,^{2}\\[2ex]
&\quad        +P_{r}\cdot E_{c}\cdot M^{2}F'\,^{2}+A^{\ast}\,F'+B^{\ast}\theta=0
    \end{split}
\\
\begin{split}
    C'' +S_{c}\left(F\,C'-C\,F'\right)-S_{c}\left(C+\frac{1}{2}\,\eta\,C'\right)+S_{c}\cdot S_{r}\,\theta''-K_{r}\cdot S_{c}=0
\end{split}
\end{align}
\end{document}

在此处输入图片描述

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