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獲得左對齊方程式

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獲得左對齊方程式

我怎麼才能將這個方程式向左對齊？

\begin{equation}
\beta_1=(n_{20}+n_{02})
\\
\beta_2={(n_{20}-n_{02})^2}+4{n_{11}}^2
\\
\beta_3={(n_{30}-3n_{12})^2}+{(3n_{21}-n_{03})^2}
\\
\beta_4={(n_{30}+n_{12})^2}+{(n_{21}-n_{03})^2}
\\
\beta_5=(n_{30}-3n_{12})(n_{30}+n_{12})[{(n_{30}+n_{12})^2}-3{(n_{21}+n_{03})^2}]+(3n_{21}-n_{03})(n_{21}+n_{03})[3{(n_{30}+n_{12})^2}-{(n_{21}+n_{03})^2}
\\
\beta_6=(n_{20}-n_{02})[{(n_{30}+n_{12})^2}-{(n_{21}+n_{03})^2}]+4n_{11}(n_{30}+n_{12})(n_{21}+n_{03})
\\
\beta_7=(3n_{21}-n_{03})(n_{30}+n_{12})[{(n_{30}+n_{12})^2}-3{(n_{21}+n_{03})^2}]-(n_{30}+3n_{12})(n_{21}+n_{03})[3{(n_{30}+n_{12})^2}-{(n_{21}+n_{03})^2}]
\label{Eq:momentsHU}
\end{equation}

答案1

它不起作用：equation適用於單行方程式。如果您只需要一個方程式編號，請使用alignedfrom amsmath, are alignedinside 。equation最長的方程式必須再次分割，你可以這樣做，例如multlined：

\documentclass[a4paper, 11pt]{book}
\usepackage[utf8]{inputenc}
\usepackage[showframe]{geometry}
\usepackage{mathtools}

\begin{document}

\begin{equation}
  \begin{aligned}
    \beta_1 & =(n_{20}+n_{02})
    \\
    \beta_2 & ={(n_{20}-n_{02})^2}+4{n_{11}}^2
    \\
    \beta_3 & ={(n_{30}-3n_{12})^2}+{(3n_{21}-n_{03})^2}
    \\
    \beta_4 & ={(n_{30}+n_{12})^2}+{(n_{21}-n_{03})^2}
    \\
    \beta_5 & =\!\begin{multlined}[t]
    (n_{30}-3n_{12})(n_{30}+n_{12})[{(n_{30}+n_{12})^2}-3{(n_{21}+n_{03})^2}]
    \\+(3n_{21}-n_{03})(n_{21}+n_{03})[3{(n_{30}+n_{12})^2}-{(n_{21}+n_{03})^2}
    \end{multlined}
    \\
    \beta_6 & =(n_{20}-n_{02})[{(n_{30}+n_{12})^2}-{(n_{21}+n_{03})^2}]+4n_{11}(n_{30}+n_{12})(n_{21}+n_{03})
    \\
    \beta_7 & =\!\begin{multlined}[t]
    (3n_{21}-n_{03})(n_{30}+n_{12})[{(n_{30}+n_{12})^2}-3{(n_{21}+n_{03})^2}]\\ -(n_{30}+3n_{12})(n_{21}+n_{03})[3{(n_{30}+n_{12})^2}-{(n_{21}+n_{03})^2}]
    \end{multlined}
    \label{Eq:momentsHU}
  \end{aligned}
\end{equation}

\end{document}

在此輸入影像描述

答案2

我只是使用align*（包含\usepackage{amsmath}在您的序言中）設置它並在適當的位置打破冗長的行。您flalign*也可以考慮使用：

在此輸入影像描述

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{align*}
  \beta_1 &= (n_{20} + n_{02}) \\
  \beta_2 &= (n_{20} - n_{02})^2 + 4 n_{11}^2 \\
  \beta_3 &= (n_{30} - 3n_{12})^2 + (3n_{21} - n_{03})^2 \\
  \beta_4 &= (n_{30} + n_{12})^2 + (n_{21} - n_{03})^2 \\
  \beta_5 &= (n_{30} - 3n_{12})(n_{30} + n_{12})[(n_{30} + n_{12})^2 - 3(n_{21} + n_{03})^2] + {} \\
          &\phantom{{}={}} \qquad (3n_{21} - n_{03})(n_{21} + n_{03})[3(n_{30} + n_{12})^2 - (n_{21} + n_{03})^2 \\
  \beta_6 &= (n_{20} - n_{02})[(n_{30} + n_{12})^2 - (n_{21} + n_{03})^2] + 4n_{11}(n_{30} + n_{12})(n_{21} + n_{03}) \\
  \beta_7 &= (3n_{21} - n_{03})(n_{30} + n_{12})[(n_{30} + n_{12})^2 - 3(n_{21} + n_{03})^2] - {} \\
          &\phantom{{}={}} \qquad (n_{30} + 3n_{12})(n_{21} + n_{03})[3(n_{30} + n_{12})^2 - (n_{21} + n_{03})^2]
\end{align*}

\end{document}

答案3

這或多或少是相同的問題方程式左對齊。您可以使用該flalign環境或載入documentclass該fleqn選項。

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