☰

是否有可能打破以下对齐？

tag-icon • tag-icon math-mode line-breaking page-breaking inline
是否有可能打破以下对齐？

我不想使用为其\allowdisplaybreak设计的显示数学环境。简而言之，是否可以alined通过定义类似的东西来打破以下内容\allowinlinebreak？

enter image description here

\documentclass{article}
\usepackage[a6paper,vmargin=15mm,hmargin=5mm]{geometry}
\usepackage{times}
\usepackage{helvet}
\usepackage{courier}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\allowdisplaybreaks
\usepackage{tcolorbox}
\usepackage{microtype}


\tcbuselibrary{breakable}
\newcounter{exercise}[subsection]
\@addtoreset{exercise}{section}


\renewcommand{\theexercise}{%
    \ifnum\value{subsection}>0\relax
        \thesubsection
    \else
        \thesection
    \fi
    .\arabic{exercise}%
}

\newenvironment{exercise}
{\par\smallskip\refstepcounter{exercise}
\begin{tcolorbox}[breakable,title=Exercise \theexercise,left=0mm,lefttitle=2mm]\ignorespaces}
{\end{tcolorbox}\par\smallskip\ignorespacesafterend}

\newif\ifinitialized
\initializedfalse
\newlength\EWD
\newlength\HWD

\def\Initializer{%
    \ifinitialized
    \else
        \initializedtrue
        \settowidth{\EWD}{${}={}$}%
        \setlength{\HWD}{0.5\dimexpr\linewidth-\EWD\relax}%
    \fi}

\AtBeginDocument{%
    \expandafter\def\expandafter\item\expandafter{%
        \expandafter\Initializer\item}%
}

\newenvironment{multi}
{\multlined[t][\HWD]}
{\endmultlined}

\begin{document}
\begin{exercise}
Simplify the following
\begin{enumerate}
    \item $\frac{3x-2}{2(x-1)}-\frac{3(x+1)}{x+2}$
\end{enumerate}
\tcblower
\begin{enumerate}
\item 
$\!
\begin{aligned}[t]
\frac{3x-2}{2(x-1)}-\frac{3(x+1)}{x+2}
    &= \begin{multi}
                \frac{3x-2}{2(x-1)}\times\frac{x+2}{x+2}\\
                {}-\frac{3(x+1)}{x+2}\times\frac{2(x-1)}{2(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{(3x-2)(x+2)}{2(x-1)(x+2)}\\
                {}-\frac{6(x+1)(x-1)}{2(x+2)(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{3x^2 +6x -2x -4}{2(x-1)(x+2)}\\
                {}-\frac{6(x^2 -x +x -1)}{2(x+2)(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{3x^2 +4x -4}{2(x-1)(x+2)}\\
                {}-\frac{6(x^2 -1)}{2(x+2)(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{3x^2 +4x -4}{2(x-1)(x+2)}\\
                {}-\frac{6x^2 -6}{2(x+2)(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{3x^2 +4x -4}{2(x-1)(x+2)}\\
                {}+\frac{-(6x^2 -6)}{2(x+2)(x-1)}
         \end{multi}\\
    &= \frac{3x^2 +4x -4 -(6x^2 -6)}{2(x-1)(x+2)}\\
    &= \frac{3x^2 +4x -4 -6x^2 +6}{2(x-1)(x+2)}\\
    &= \frac{-3x^2 +4x +2}{2(x-1)(x+2)}
\end{aligned}
$
\end{enumerate}
\end{exercise}
\end{document}

答案1

使用内联数学所以这样做是错误的（但每次你发例子时我都会说同样的话）。但是如果你必须这样做，那么你必须……

enter image description here

\documentclass{article}
\usepackage[a6paper,vmargin=15mm,hmargin=5mm]{geometry}
\usepackage{times}
\usepackage{helvet}
\usepackage{courier}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\allowdisplaybreaks
\usepackage{tcolorbox}
\usepackage{microtype}


\tcbuselibrary{breakable}
\newcounter{exercise}[subsection]
\@addtoreset{exercise}{section}


\renewcommand{\theexercise}{%
    \ifnum\value{subsection}>0\relax
        \thesubsection
    \else
        \thesection
    \fi
    .\arabic{exercise}%
}

\newenvironment{exercise}
{\par\smallskip\refstepcounter{exercise}
\begin{tcolorbox}[breakable,title=Exercise \theexercise,left=0mm,lefttitle=2mm]\ignorespaces}
{\end{tcolorbox}\par\smallskip\ignorespacesafterend}

\newif\ifinitialized
\initializedfalse
\newlength\EWD
\newlength\HWD

\def\Initializer{%
    \ifinitialized
    \else
        \initializedtrue
        \settowidth{\EWD}{${}={}$}%
        \setlength{\HWD}{0.5\dimexpr\linewidth-\EWD\relax}%
    \fi}

\AtBeginDocument{%
    \expandafter\def\expandafter\item\expandafter{%
        \expandafter\Initializer\item}%
}

\newenvironment{multi}
{\multlined[t][\HWD]}
{\endmultlined}

\begin{document}
\begin{exercise}
Simplify the following
\begin{enumerate}
    \item $\frac{3x-2}{2(x-1)}-\frac{3(x+1)}{x+2}$
\end{enumerate}
\tcblower
\begin{enumerate}
\item 
\setbox0\vbox{\rightskip\fill$\break\begin{aligned}[t]
\frac{3x-2}{2(x-1)}-\frac{3(x+1)}{x+2}
    &= \begin{multi}
                \frac{3x-2}{2(x-1)}\times\frac{x+2}{x+2}\\
                {}-\frac{3(x+1)}{x+2}\times\frac{2(x-1)}{2(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{(3x-2)(x+2)}{2(x-1)(x+2)}\\
                {}-\frac{6(x+1)(x-1)}{2(x+2)(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{3x^2 +6x -2x -4}{2(x-1)(x+2)}\\
                {}-\frac{6(x^2 -x +x -1)}{2(x+2)(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{3x^2 +4x -4}{2(x-1)(x+2)}\\
                {}-\frac{6(x^2 -1)}{2(x+2)(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{3x^2 +4x -4}{2(x-1)(x+2)}\\
                {}-\frac{6x^2 -6}{2(x+2)(x-1)}
         \end{multi}\\
    &= \begin{multi}
                \frac{3x^2 +4x -4}{2(x-1)(x+2)}\\
                {}+\frac{-(6x^2 -6)}{2(x+2)(x-1)}
         \end{multi}\\
    &= \frac{3x^2 +4x -4 -(6x^2 -6)}{2(x-1)(x+2)}\\
    &= \frac{3x^2 +4x -4 -6x^2 +6}{2(x-1)(x+2)}\\
    &= \frac{-3x^2 +4x +2}{2(x-1)(x+2)}
\end{aligned}\break$\par
\unskip\setbox0\lastbox\unskip\unpenalty
\setbox0\lastbox
\setbox0\hbox{\unhbox0\unskip\unpenalty\global\setbox1\lastbox}}%
\unvbox1
\end{enumerate}
\end{exercise}
\end{document}

相关内容