答案1
listliketab
用enumerate 代替的解决方案:
\documentclass[11pt, a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lipsum}
\usepackage{listliketab}
\begin{document}
\storestyleof{enumerate}
\newcounter{tabenum}\setcounter{tabenum}{0}
\newcommand{\tabitem}{\refstepcounter{tabenum}\thetabenum)}
\setlength{\tabcolsep}{1em}
\begin{listliketab}
\begin{tabular}{L*{2}{>{$}l<{$}}}
\tabitem & a = 5, & b = 3 \,; \\
\tabitem & a = 4, & b = 6 \,; \\
\tabitem & a =3{,}2, & b = 2{,}3 \,; \\
\tabitem & a = 2\sqrt{5}, & b = 3\sqrt{5} \,; \\
\tabitem & a = \sqrt{3}, & b = \sqrt{7} \,; \\
\tabitem & a = 1, & b = √{0{,}1}.
\end{tabular}
\end{listliketab}
\end{document}
另一个解决方案基于@Werner 的代码在同一文档中切换左右方程编号以及fleqn
来自 的环境nccmath
。=
符号将在两列中对齐:
\documentclass[11pt, a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lipsum}
\usepackage{amsmath}
\usepackage{nccmath}
\makeatletter
\newcommand{\leqnomode}{\tagsleft@true}
\newcommand{\reqnomode}{\tagsleft@false}
\makeatother
\begin{document}
\begin{fleqn}[2em]
\leqnomode
\begin{alignat}{2}
a &= 5, & b & = 3 \,; \\
a & = 4, & b & = 6 \,; \\
a & =3{,}2, & b & = 2{,}3 \, ; \\
a & = 2\sqrt{5}, &\qquad b & = 3\sqrt{5} \, ; \\
a & = \sqrt{3}, & b & = \sqrt{7} \, ; \\
a & = 1, & b & = \sqrt{0{,}1}.
\end{alignat}
\end{fleqn}
\begin{align}
f(x) &= ax^2 + bx + c \\
g(x) &= dx^2 + ex + f
\end{align}
\end{document}