有没有办法将并排放置的 warwidth 数组顶部对齐？
\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[italian]{babel}
\usepackage[
a4paper,
margin=15mm,
bindingoffset=2mm,
heightrounded,
]{geometry}
\usepackage{amsmath}
\usepackage{microtype}
\usepackage{xcolor}
\usepackage{varwidth}
\usepackage{enumitem}
\usepackage{hyperref}
\definecolor{lightgray}{HTML}{EFEFEF}
\DeclareMathOperator{\sen}{sen}
\begin{document}
\begingroup\setlength{\fboxsep}{2pt}
\colorbox{white}{%
\begin{varwidth}{\textwidth}
POTENZE
\begin{itemize}[leftmargin=*]
\item $a^{1} = a$
\item \textcolor{blue}{$a^{0} = 1$}
\item $a^{n} \cdot a^{m} = a^{n+m}$
\item \textcolor{blue}{$\dfrac{a^{n}}{a^{m}} = a^{n-m}$}
\item $ a^{-n} = \dfrac{1}{a^{n}}$
\item \textcolor{blue}{${(a^{n})}^{m} = a^{n \cdot m}$}
\item $a^{n} \cdot b^{n} = {(a \cdot b)}^{n}$
\item \textcolor{blue}{$\dfrac{a^{n}}{b^{n}} = \left(\dfrac{a}{b}\right)^{n}, b \neq 0$}
\item $a^{ \tfrac{m}{n} } = \sqrt[n]{a^{m}}$
\end{itemize}
\end{varwidth}%
}\endgroup
\quad
\begingroup\setlength{\fboxsep}{6pt}
\colorbox{lightgray}{%
\begin{varwidth}{\textwidth}
LOGARITMI
\begin{itemize}[leftmargin=*]
\item $a^{\log_{a}b} = b$
\item \textcolor{blue}{$\log_{a}{bc} = \log_{a}{b} + \log_{a}{c}$}
\item $\log_{a}{\dfrac{b}{c}} = \log_{a}{b} - \log_{a}{c}$
\item \textcolor{blue}{$c \log_{a} b = log_{a} b^{c}$}
\item $\log_{a} b = \dfrac{1}{\log_{b} a}$
\end{itemize}
\end{varwidth}%
}\endgroup
\quad
\begingroup\setlength{\fboxsep}{6pt}
\colorbox{lightgray}{%
\begin{varwidth}{\textwidth}
DERIVATE
\begin{itemize}[leftmargin=*]
\item $\dfrac{d}{dx}a= 0$
\item \textcolor{blue}{$\dfrac{d}{dx}x = 1$}
\item $\dfrac{d}{dx}ax^n = anx^{n-1}$
\item \textcolor{blue}{$\dfrac{d}{dx}\ln x = \dfrac{1}{x}$, $x > 0$}
\item $\dfrac{d}{dx}\log_{a}x = \dfrac{1}{x\ln a} = \dfrac{\log_{a}e}{x}$, $x > 0$
\item \textcolor{blue}{$\dfrac{d}{dx}\sen x = \cos x$}
\item $\dfrac{d}{dx}\cos x = -\sen x$
\end{itemize}
\end{varwidth}%
}\endgroup
\quad
\end{document}
我需要将中间的一个对齐到顶部
答案1
只需使用可选参数[t]：
\begin{varwidth}[t]{\textwidth}
