我想将以下方程式放在文本的左侧，但我不知道哪个命令对此有用。我使用了对齐，但它将所有方程式都放在了纸的最右侧。更准确地说，我想将所有方程式放在彼此下方，并且它们的位置在纸的左侧。
\begin{document}
\begin{align*}
(f,f)_{[x^3]_{1}} = (f,f)_{[x^3]_{1}}=0 \\
(f,g)_{[x^3]_{1}} =(f,g)_{[x^3]_{3}}=0\\
(f,h)_{[xxy]_{1}}=(f,h)_{[xxy]_{3}}=0\\
(f,p)_{[xxy]_{1}}=(f,p)_{[xxy]_{3}}=0\\
(f,m)_{[xxt]_{1}}=[xx]_{1} \dashv t - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
(f,m)_{[xxt]_{3}}=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
(g,f)_{[x^3]_{1}}=(g,f)_{[x^3]_{3}}=0\\
\end{align*}
\end{document}
答案1
该包nncmath提供fleqn将方程式推送至文本左边框的环境：
\documentclass{article}
\usepackage{nccmath}
%---------------- show page layout. don't use in a real document!
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%
\begin{document}
\begin{fleqn}
\begin{gather*}
(f,f)_{[x^3]_{1}} = (f,f)_{[x^3]_{1}}=0 \\
(f,g)_{[x^3]_{1}} =(f,g)_{[x^3]_{3}}=0 \\
(f,h)_{[xxy]_{1}}=(f,h)_{[xxy]_{3}}=0 \\
(f,p)_{[xxy]_{1}}=(f,p)_{[xxy]_{3}}=0 \\
(f,m)_{[xxt]_{1}}=[xx]_{1} \dashv t - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
(f,m)_{[xxt]_{3}}=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
(g,f)_{[x^3]_{1}}=(g,f)_{[x^3]_{3}}=0\\
\end{gather*}
\end{fleqn}
\end{document}
（红线表示文本边框）
答案2
以下是一个可能的建议，包括三种不同的对齐方式（红线表示边距）：
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{flalign*}
(f,f)_{[x^3]_{1}} &= (f,f)_{[x^3]_{1}}=0 &&\\
(f,g)_{[x^3]_{1}} &=(f,g)_{[x^3]_{3}}=0 \\
(f,h)_{[xxy]_{1}}&=(f,h)_{[xxy]_{3}}=0 \\
(f,p)_{[xxy]_{1}}&=(f,p)_{[xxy]_{3}}=0 \\
(f,m)_{[xxt]_{1}}&=[xx]_{1} \dashv t - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
(f,m)_{[xxt]_{3}}&=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
(g,f)_{[x^3]_{1}}&=(g,f)_{[x^3]_{3}}=0\\
\end{flalign*}
\begin{align*}
(f,f)_{[x^3]_{1}} &= (f,f)_{[x^3]_{1}}=0 \\
(f,g)_{[x^3]_{1}} &=(f,g)_{[x^3]_{3}}=0 \\
(f,h)_{[xxy]_{1}}&=(f,h)_{[xxy]_{3}}=0 \\
(f,p)_{[xxy]_{1}}&=(f,p)_{[xxy]_{3}}=0 \\
(f,m)_{[xxt]_{1}}&=[xx]_{1} \dashv t - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
(f,m)_{[xxt]_{3}}&=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
(g,f)_{[x^3]_{1}}&=(g,f)_{[x^3]_{3}}=0\\
\end{align*}
\begin{flalign*}
&(f,f)_{[x^3]_{1}} = (f,f)_{[x^3]_{1}}=0 &&\\
&(f,g)_{[x^3]_{1}} =(f,g)_{[x^3]_{3}}=0 \\
&(f,h)_{[xxy]_{1}}=(f,h)_{[xxy]_{3}}=0 \\
&(f,p)_{[xxy]_{1}}=(f,p)_{[xxy]_{3}}=0 \\
&(f,m)_{[xxt]_{1}}=[xx]_{1} \dashv t - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
&(f,m)_{[xxt]_{3}}=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
&(g,f)_{[x^3]_{1}}=(g,f)_{[x^3]_{3}}=0\\
\end{flalign*}
\end{document}
答案3
只需添加一些小变体，其中一些方程式居中而另一些方程式左对齐，当然不如@leandriis 答案那么优雅，但它可能会有用：
\documentclass{article}
\usepackage{amsmath}
\usepackage{blindtext}
\begin{document}
\begin{align*}
& (f,f)_{[x^3]_{1}} = (f,f)_{[x^3]_{1}}=0 \\
& (f,g)_{[x^3]_{1}} = (f,g)_{[x^3]_{3}}=0 \\
& (f,h)_{[xxy]_{1}} = (f,h)_{[xxy]_{3}}=0 \\
& (f,p)_{[xxy]_{1}} = (f,p)_{[xxy]_{3}}=0 \\
& (f,m)_{[xxt]_{1}} = [xx]_{1} \dashv t - x \dashv ([xt]_{1} - [tx]_{2}) = [xtx]_{1}\\
& (f,m)_{[xxt]_{3}} = [xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
& (g,f)_{[x^3]_{1}} =(g,f)_{[x^3]_{3}}=0\\
\end{align*}
%some blind text to check that margins are correct
\blindtext
\begin{align*}
& (f,f)_{[x^3]_{1}} = (f,f)_{[x^3]_{1}}=0 \\
& (f,g)_{[x^3]_{1}} = (f,g)_{[x^3]_{3}}=0 \\
\noalign{\hspace{-\parindent}$(f,h)_{[xxy]_{1}} = (f,h)_{[xxy]_{3}}=0$}
\noalign{\hspace{-\parindent}$(f,p)_{[xxy]_{1}} = (f,p)_{[xxy]_{3}}=0$}
& (f,m)_{[xxt]_{1}} = [xx]_{1} \dashv t - x \dashv ([xt]_{1} - [tx]_{2}) = [xtx]_{1}\\
& (f,m)_{[xxt]_{3}} = [xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
& (g,f)_{[x^3]_{1}} =(g,f)_{[x^3]_{3}}=0\\
\end{align*}
\end{document}
输出为：
