长评论管理

Question

我希望将评论保留在侧面，因此我不想使用互文或短互文，而是将长行设为两行，并保持所有内容等距！有什么技巧/命令？

\documentclass{article}
\usepackage{amsmath,xcolor,geometry}
\begin{document}
$\begin{aligned}[t] \dfrac{20-5x^{2}}{x^{2}+x-6}&=\dfrac{5(4-x^{2})}{(x+3) 
(x-2)} && \textcolor{blue}{\text{Factor}} \\
                                                   &=\dfrac{5(2+x)(2-x)} 
{(x+3)(x-2)} && \textcolor{blue}{\text{Factor completely}} \\

&=\dfrac{5(2+x) \cdot \textcolor{red}{-1}(\colorbox{cyan!25!white}{$x-2$})} 
{(x+3)(\colorbox{cyan!25!white}{$x-2$})} && \textcolor{blue}{\text{Notice 
opposites in $2-x$ and $x-2$. Write $2-x$ as $-1(x-2)$ and simplify}} \\

&=-\dfrac{5(2+x)}{x+3}
\end{aligned}$
\end{document}

Answer 1

在这种情况下，我会采取以下措施。

\documentclass{article}

\usepackage{amsmath,xcolor,geometry}

\newbox\aebox
\newcommand\aecomment[1]{%%
  \begin{lrbox}\aebox
    \begin{minipage}[t]{2in}
      \raggedright
      \textcolor{blue}{#1}%%
    \end{minipage}%%
  \end{lrbox}%%
  \raisebox{0pt}[\height][0pt]{\usebox{\aebox}}}
\begin{document}

$
  \begin{aligned}[t] 
    \dfrac{20-5x^{2}}{x^{2}+x-6} &=\dfrac{5(4-x^{2})}{(x+3) (x-2)} && \textcolor{blue}{\text{Factor}}            \\
                                 &=\dfrac{5(2+x)(2-x)}{(x+3)(x-2)} && \textcolor{blue}{\text{Factor completely}} \\
                                 &=\dfrac{5(2+x) \cdot \textcolor{red}{-1}(\colorbox{cyan!25!white}{$x-2$})} 
                                         {(x+3)(\colorbox{cyan!25!white}{$x-2$})} 
                                 %% && \textcolor{blue}{\text{Notice opposites in $2-x$ and $x-2$. Write $2-x$ as $-1(x-2)$ and simplify}} \\
                                 && \aecomment{Notice opposites in $2-x$ and $x-2$. Write $2-x$ as $-1(x-2)$ and simplify} \\
                                 &=-\dfrac{5(2+x)}{x+3}
  \end{aligned}
$

\end{document}

现在，我应该提醒一下，我已经消除了评论的深度。这意味着如果你写了类似的东西：

$
  \begin{aligned}[t] 
    \dfrac{20-5x^{2}}{x^{2}+x-6} &=\dfrac{5(4-x^{2})}{(x+3) (x-2)} && \textcolor{blue}{\text{Factor}}            \\
                                 &=\dfrac{5(2+x)(2-x)}{(x+3)(x-2)} && \textcolor{blue}{\text{Factor completely}} \\
                                 &=\dfrac{5(2+x) \cdot \textcolor{red}{-1}(\colorbox{cyan!25!white}{$x-2$})} 
                                         {(x+3)(\colorbox{cyan!25!white}{$x-2$})} 
                                 %% && \textcolor{blue}{\text{Notice opposites in $2-x$ and $x-2$. Write $2-x$ as $-1(x-2)$ and simplify}} \\
                                 && \aecomment{Notice opposites in $2-x$ and $x-2$. Write $2-x$ as $-1(x-2)$ and simplify} \\
                                 &=-\dfrac{5(2+x)}{x+3}
                                 && \textcolor{red}{\text{\bfseries Smashes into line above!!!}}
  \end{aligned}
$

您将遇到评论冲突。因此，请注意。

我倾向于谨慎使用此类注释。因此，此类冲突对我来说不是一个重大问题。但在这里您可以看到如果我不将深度置零会发生什么：

如果您不介意长注释在等式中产生的额外空间，那么只需删除部分\raisebox细节即可。

这种崩溃可能是避免\raisebox通过实际使用增加框按指定量移动。在这里，我重新定义了\aecomment

\newbox\aebox
\newcommand\aecomment[2][0pt]{%%
  \begin{lrbox}\aebox
    \begin{minipage}[t]{2in}
      \raggedright
      \textcolor{blue}{#2}%%
    \end{minipage}%%
  \end{lrbox}%%
  \raisebox{#1}[\height][0pt]{\usebox{\aebox}}}

调用方式如下：

$
  \begin{aligned}[t] 
    \dfrac{20-5x^{2}}{x^{2}+x-6} &=\dfrac{5(4-x^{2})}{(x+3) (x-2)} && \textcolor{blue}{\text{Factor}}            \\
                                 &=\dfrac{5(2+x)(2-x)}{(x+3)(x-2)} && \textcolor{blue}{\text{Factor completely}} \\
                                 &=\dfrac{5(2+x) \cdot \textcolor{red}{-1}(\colorbox{cyan!25!white}{$x-2$})} 
                                         {(x+3)(\colorbox{cyan!25!white}{$x-2$})} 
                                 && \aecomment[2.5ex]{Notice opposites in $2-x$ and $x-2$. Write $2-x$ as $-1(x-2)$ and simplify} \\
                                 &=-\dfrac{5(2+x)}{x+3}
                                 && \textcolor{red}{\text{\bfseries No crashing into line above!!!}}
  \end{aligned}
$

不会发生冲突，而且也不会影响方程本身的间距。

Answer 2

使用软件包的简单解决方案stackengine：

\documentclass{article}
\usepackage{amsmath,xcolor,geometry}
\usepackage[usestackEOL]{stackengine}

\begin{document}

$\begin{aligned}[t] \dfrac{20-5x^{2}}{x^{2}+x-6}&=\dfrac{5(4-x^{2})}{(x+3)
(x-2)} && \textcolor{blue}{\text{Factor}} \\
                                                   &=\dfrac{5(2+x)(2-x)}
{(x+3)(x-2)} && \textcolor{blue}{\text{Factor completely}} \\
&=\dfrac{5(2+x) \cdot \textcolor{red}{-1}(\colorbox{cyan!25!white}{$x-2$})}
{(x+3)(\colorbox{cyan!25!white}{$x-2$})} && \textcolor{blue}{\Centerstack[l]{Notice
opposites in $2-x$ and $x-2$. \\Write $2-x$ as $-1(x-2)$ and simplify}} \\
&=-\dfrac{5(2+x)}{x+3}
\end{aligned}$

\end{document}