我想扩大投影仪块底部方程式后的空间，使其看起来更平滑。
我尝试使用\vspace{1in}、\leavevmode\hphantom{ }和\vspace{0.1in}。但它会产生较长的垂直空间
\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{multicol}
\usepackage{pifont}
\usepackage{tasks}
\usepackage{empheq}
\usepackage{tikz,tkz-tab}
\usetikzlibrary{calc,positioning}
\usepackage{multirow}
\usepackage{makecell}
\newcommand{\abs}[1]{\left\lvert#1\right\rvert} % Commande pour obtenir la valeur absolue.
\DeclarePairedDelimiter{\openintvl}{]}{[}
\renewcommand{\arraystretch}{2}
\usetheme{Madrid}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\begin{frame}{Differentiability }
Determine whether each of the following functions is (a) continuous, and (b) differentiable.
\begin{block}{Exercise}
\begin{multicols}{2}
\begin{enumerate}\rightskip\leftm
\item $f(x)=2x-5$
\item $f(x)=x^{2}+\dfrac{1}{2}x+1$
\item $f(x)=(2x+1)(x^{2}+1)$
\item $f(x)=\sin x\cos x$
\item $f(x)=\dfrac{1}{x^{2}}$
\item $f(x)=\dfrac{1}{2+\cos x}$
\item $f(x)=\dfrac{x^{2}+7}{x+1}$
\item $f(x)=\dfrac{\sin x}{x}$
\item $f(x)=\sqrt{x+2}$
\item $f(x)=\sqrt{x^{2}+1}$
\item $f(x)=\dfrac{1}{\sqrt{x-1}}$
\item $f(x)=\dfrac{x}{\sqrt{x^{2}+1}}$
\item $f(x)=\cos(2x)$
\item $f(x)=\sin\left(3x+\dfrac{\pi}{4}\right)$
\end{enumerate}
\end{multicols}
\end{block}
\end{frame}
\end{document}
答案1
这里有两个快速但或多或少有点肮脏的解决方案。
第一种情况，很脏，加一个隐 item在每列的底部。第二种情况，稍微不那么脏，但有点令人惊讶，在\vspace{0pt}后面添加一个\end{multicols}
\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{multicol}
\usepackage{pifont}
\usepackage{tasks}
\usepackage{empheq}
\usepackage{tikz,tkz-tab}
\usetikzlibrary{calc,positioning}
\usepackage{multirow}
\usepackage{makecell}
\newcommand{\abs}[1]{\left\lvert#1\right\rvert} % Commande pour obtenir la valeur absolue.
\DeclarePairedDelimiter{\openintvl}{]}{[}
\renewcommand{\arraystretch}{2}
\usetheme{Madrid}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\begin{frame}{Differentiability }
Determine whether each of the following functions is (a) continuous, and (b) differentiable.
\begin{block}{Exercise}
\begin{multicols}{2}
\begin{enumerate}%\rightskip\leftm
\item $f(x)=2x-5$
\item $f(x)=x^{2}+\dfrac{1}{2}x+1$
\item $f(x)=(2x+1)(x^{2}+1)$
\item $f(x)=\sin x\cos x$
\item $f(x)=\dfrac{1}{x^{2}}$
\item $f(x)=\dfrac{1}{2+\cos x}$
\item $f(x)=\dfrac{x^{2}+7}{x+1}$
\item[]
\item $f(x)=\dfrac{\sin x}{x}$
\item $f(x)=\sqrt{x+2}$
\item $f(x)=\sqrt{x^{2}+1}$
\item $f(x)=\dfrac{1}{\sqrt{x-1}}$
\item $f(x)=\dfrac{x}{\sqrt{x^{2}+1}}$
\item $f(x)=\cos(2x)$
\item $f(x)=\sin\left(3x+\dfrac{\pi}{4}\right)$
\item[]
\end{enumerate}
\end{multicols}
\end{block}
\end{frame}
\begin{frame}{Differentiability }
Determine whether each of the following functions is (a) continuous, and (b) differentiable.
\begin{block}{Exercise}
\begin{multicols}{2}
\begin{enumerate}%\rightskip\leftm
\item $f(x)=2x-5$
\item $f(x)=x^{2}+\dfrac{1}{2}x+1$
\item $f(x)=(2x+1)(x^{2}+1)$
\item $f(x)=\sin x\cos x$
\item $f(x)=\dfrac{1}{x^{2}}$
\item $f(x)=\dfrac{1}{2+\cos x}$
\item $f(x)=\dfrac{x^{2}+7}{x+1}$
\item $f(x)=\dfrac{\sin x}{x}$
\item $f(x)=\sqrt{x+2}$
\item $f(x)=\sqrt{x^{2}+1}$
\item $f(x)=\dfrac{1}{\sqrt{x-1}}$
\item $f(x)=\dfrac{x}{\sqrt{x^{2}+1}}$
\item $f(x)=\cos(2x)$
\item $f(x)=\sin\left(3x+\dfrac{\pi}{4}\right)$
\end{enumerate}
\end{multicols}
\vspace{0pt}
\end{block}
\end{frame}
\end{document}
答案2
我认为该multicols软件包与 Beamer 不完全兼容。列平衡的概念很难应用于 Beamer 的框架概念。
我尝试设计一个更“类似 beamer”的解决方案，使用columns环境进行两列布局。当然，您必须放弃 multicols 包的一些好处，例如自动分列。
为了使它至少看起来像一个连续的列，我定义了两个新的枚举环境，将项目计数器从一个传递到另一个。
\documentclass{beamer}
\usetheme{Madrid}
\newcounter{ContinueEnumerate}
\newenvironment{BeginEnumerate}{
\begin{enumerate}%\rightskip\leftm
}{
\setcounter{ContinueEnumerate}{\arabic{enumi}}
\end{enumerate}
}
\newenvironment{ContinueEnumerate}{
\begin{enumerate}
\setcounter{enumi}{\arabic{ContinueEnumerate}}
}{
\end{enumerate}
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\begin{frame}{Differentiability }
Determine whether each of the following functions is (a) continuous, and (b) differentiable.
\begin{block}{Exercise}
\begin{columns}[t]
\column{0.5\linewidth}
\begin{BeginEnumerate}
\item $f(x) = 2x-5$
\item $f(x) = x^{2}+\dfrac{1}{2}x+1$
\item $f(x) = (2x+1)(x^{2}+1)$
\item $f(x) = \sin x\cos x$
\item $f(x) = \dfrac{1}{x^{2}}$
\item $f(x) = \dfrac{1}{2+\cos x}$
\item \label{x} $f(x) = \dfrac{x^{2}+7}{x+1}$
\end{BeginEnumerate}
\column{0.5\linewidth}
\begin{ContinueEnumerate}
\item $f(x) = \dfrac{\sin x}{x}$
\item $f(x) = \sqrt{x+2}$
\item $f(x) = \sqrt{x^{2}+1}$
\item $f(x) = \dfrac{1}{\sqrt{x-1}}$
\item $f(x) = \dfrac{x}{\sqrt{x^{2}+1}}$
\item $f(x) = \cos(2x)$
\item $f(x) = \sin\left(3x+\dfrac{\pi}{4}\right)$
\end{ContinueEnumerate}
\end{columns}
\end{block}
\end{frame}
\end{document}