我想让 LHS 上的拆分方程左对齐。我的 2 个尝试如下所示,即选项 A 和选项 B。请参阅源代码了解详细信息。
\documentclass[10pt]{article}
\usepackage[a6paper,vmargin=15mm,hmargin=5mm]{geometry}
\usepackage{mathtools}
\begin{document}
\begin{enumerate}
\item Hey baby!
\item
$\!
\begin{aligned}[t]
2-3x(x-1)\\
{}-3(x-2y)(x+2y)
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3(x^2 +2xy -2xy -4y^2)
\end{multlined}\\
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3(x^2 -4y^2)
\end{multlined}\\
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3x^2 +12y^2
\end{multlined}\\
&= -6x^2 +3x +12y^2 +2
\end{aligned}
$
\item
$\!
\begin{aligned}[t]
\begin{multlined}[t][4cm]
2-3x(x-1)\\
{}-3(x-2y)(x+2y)
\end{multlined}\\
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3(x^2 +2xy -2xy -4y^2)
\end{multlined}\\
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3(x^2 -4y^2)
\end{multlined}\\
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3x^2 +12y^2
\end{multlined}\\
&= -6x^2 +3x +12y^2 +2
\end{aligned}
$
\end{enumerate}
\end{document}
如何解决这个问题呢?
答案1
这将为您提供选项 b:
\documentclass[10pt]{article}
\usepackage[a6paper,vmargin=15mm,hmargin=5mm]{geometry}
\usepackage{mathtools}
\begin{document}
\begin{enumerate}
\setcounter{enumi}{2}
\item
$
\begin{multlined}[t]
2-3x(x-1)\\
\begin{aligned}
{}-3(x-2y)(x+2y)
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3(x^2 +2xy -2xy -4y^2)
\end{multlined}\\
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3(x^2 -4y^2)
\end{multlined}\\
&= \!
\begin{multlined}[t][4cm]
2 -3x^2 +3x\\
{}-3x^2 +12y^2
\end{multlined}\\
&= -6x^2 +3x +12y^2 +2
\end{aligned}
\end{multlined}
$
\end{enumerate}
\end{document}