多行乳膠支架

Question

如何寫出圖中所示左右兩側帶有大括號的方程式？感謝您的幫忙！

Answer 1

這是有效的，因為stackengine可以設定具有固定基線跳躍的方程式。

\documentclass{article}
\usepackage{tabstackengine,amsmath}
\TABstackMath
\stackMath
\begin{document}
\[
%\setstackgap{L}{1.1\baselineskip}% VERTICAL SPACING OF CENTER EQUATIONS
%\renewcommand\stacktype{L}% GIVES LEFT/RIGHT EQS SAME VERTICAL SPACING AS CENTER EQS
\Centerstack{
\\
\\
\alignstackanchor{a=&b}{a=&b}{\left\{\begin{aligned}\\\\\end{aligned}\right.}\\
\\
\\
\alignstackanchor{a=&b}{a=&b}{\left\{\begin{aligned}\\\\\end{aligned}\right.}\\
\\
}
\alignCenterstack{
a=&b\\
a=&b\\
a=&b\\
a=&b\\
a=&b\\
a=&b\\
a=&b\\
a=&b
}
\Centerstack{
\smash{\phantom{\left.\begin{aligned}\\\\\end{aligned}\right\}}}a = b\\
\\
\\
{\left.\begin{aligned}\\\\\end{aligned}\right\}}\alignstackanchor{a=&b}{a=&b}\\
\\
\\
{\left.\begin{aligned}\\\\\end{aligned}\right\}}\alignstackanchor{a=&b}{a=&b}\\
}
\]
\end{document}

Answer 2

解決方案bigdelim：

\documentclass{article}
\usepackage{amsmath}
\usepackage{array, bigdelim} 

\begin{document}


\[\setlength{\extrarowheight}{3pt} \begin{array}{c@{\,}c@{\enspace}c }
 & a = b & a = b \\
 \ldelim\{{3}{*}[$ \begin{gathered}
 a = b \\ a = b
\end{gathered} $ ] & a = b \\
& a = b & \rdelim\}{3}{*} [$ \begin{gathered}
 a = b \\ a = b
\end{gathered} $ ]\\
  & a = b \\
\ldelim\{{3}{*}[$ \begin{gathered}
a = b \\ a = b
\end{gathered} $ ] & a = b \\
  & a = b \\
  & a = b \\
  & a = b \\
\end{array} \]%

\end{document} 

Answer 3

只是一個小小的嘗試，但Steven B. Segletes答案比我的好 100 倍，這是MWE我的嘗試：

\documentclass{book}
\usepackage{mathtools,multirow}
\begin{document}

\begin{tabular}{ccc}
&$\begin{matrix}
a=b\\
a=b
\end{matrix}$ &$a=b$\\
$\begin{matrix}
a=b\\
a=b
\end{matrix}$ &\multirow{-2}{*}{\Bigg\{} $\begin{matrix}
a=b\\
a=b
\end{matrix}$ &\multirow{2}{*}{\Bigg\}} $\begin{matrix}
a=b\\
a=b
\end{matrix}$\\[19pt]
$\begin{matrix}
a=b\\
a=b
\end{matrix}$ &\multirow{-2}{*}{\Bigg\{} $\begin{matrix}
a=b\\
a=b\\
a=b\\
a=b
\end{matrix}$ &\multirow{2}{*}{\Bigg\}} $\begin{matrix}
a=b\\
a=b
\end{matrix}$
\end{tabular}

\end{document}

Answer 4

這是一個解決{NiceArray}方案nicematrix。

\documentclass{article}
\usepackage{amsmath}
\usepackage{nicematrix}

\begin{document}

\[\setlength{\extrarowheight}{3pt} 
\begin{NiceArray}{c@{\hspace{1.5em}}c@{\hspace{1.5em}}c}
& a=b & a=b \\
\Block{3-1}{\begin{gathered}a=b\\ a=b\end{gathered}}
& a=b \\
& a=b & \Block{3-1}{\begin{gathered}a=b\\ a=b\end{gathered}}\\
& a=b \\
\Block{3-1}{\begin{gathered}a=b\\ a=b\end{gathered}}
& a=b \\
& a=b \\
& a=b \\
& a=b \\
\CodeAfter
  \SubMatrix{\lbrace}{2-2}{4-2}{.}
  \SubMatrix{.}{3-2}{5-2}{\rbrace}
  \SubMatrix{\lbrace}{5-2}{7-2}{.}
\end{NiceArray} \]%

\end{document}