我收到了 parbox 中缺失}或丢失的多个错误$，其中有两行，但我不明白为什么......我已经整理了一个 MWE 来准确复制问题所在。
\documentclass{standalone}
\usepackage{color}
\def\df{\rule{0pt}{1.8ex}}
\begin{document}
\Large
\raisebox{+20.0pt}{%
\fboxrule=2pt%
\fcolorbox{blue}{white}{%
\parbox[t][4.0cm]{8.0cm}{%
\addvspace{0.5cm}\centering
{\boldmath\color{blue}
$\displaystyle
{x} \;=\; -\, \frac{1}{k \,-\, 1}\,x_{\df{1}} \,+\,
\frac{k}{k \,-\, 1}\,x_{\df{2}}
$}\\[0.5cm]
{\boldmath\color{blue}
$\displaystyle
{y} \;=\; -\, \frac{1}{k \,-\, 1 }\,y_{\df{1}} \,+\,
\frac{k}{ k \,-\, 1 }\,y_{\df{2}}
$}%
}%
}%
}%end raisebox
\;\;\textbf{.}
\end{document}
错误如下：
! Missing $ inserted.
<inserted text>
$
l.23 }
%end raisebox
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Extra }, or forgotten $.
<recently read> }
l.23 }
%end raisebox
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Missing $ inserted.
<inserted text>
$
l.23 }
%end raisebox
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Missing } inserted.
<inserted text>
}
l.23 }
%end raisebox
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
答案1
您有太多的手动标记：很少需要\;，\,有时很有用，但不适合您的用例。
使用aligned提供的amsmath。我还添加了\bfrac与减号粗细相匹配的分数线。为了降低下标，请使用空上标。
这个奇怪的border设置是为了解决似乎是一个错误的现象standalone；7pt 是半\fboxsep加\fboxrule。
\documentclass[border={-7pt 0 0 0}]{standalone}
\usepackage{amsmath,color}
\newcommand{\bfrac}[2]{%
\genfrac{}{}{0.6pt}{}{#1}{#2}%
}
\begin{document}
\setlength{\fboxrule}{2pt}
\setlength{\fboxsep}{10pt}
\fcolorbox{blue}{white}{%
\boldmath\color{blue}%
$\!\begin{aligned}
x &= - \bfrac{1}{k-1}x^{}_{1} + \bfrac{k}{k-1}x^{}_{2}
\\
y &= - \bfrac{1}{k-1}y^{}_{1} + \bfrac{k}{k-1}y^{}_{2}
\end{aligned}$%
}
\end{document}
答案2
使用默认数学表达式\[ ... \]：
\documentclass{standalone}
\usepackage{mathtools}
\usepackage{xcolor}
\usepackage{varwidth}
\begin{document}
{\Large\fboxrule=2pt\fboxsep=10pt
\fcolorbox{blue}{white}{%
\begin{varwidth}{\linewidth}
\[ x = -\frac{ 1 }{ k - 1 }\,x_1 + \frac{ k }{ k - 1 }\,x_2 \]
\[ y = -\frac{ 1 }{ k - 1 }\,y_1 + \frac{ k }{ k - 1 }\,y_2 \]
\end{varwidth}}%
}
\end{document}
