Source
%@Auteur:Véronique Glaçon\par
Après avoir découpé les dominos ci-dessous, place les pour former un parcours rectangulaire
en respectant la règle suivante : {\em deux côtés qui sont adjacents
doivent être égaux}.
\hspace{0.5cm}

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\begin{table}[h!]
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$2(x+7)$ \\ \hspace{1cm}} & \centering{$4x+3-7x+9$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular} \end{minipage}
\hfill
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$2x(4-3x)$ \\ \hspace{1cm}} & \centering{$3(x-1)$ \\ \hspace{1cm}} \tabularnewline \hline
\end{tabular}
\end{minipage}\end{table}

\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$3-2x+4+4x$ \\ \hspace{1cm}} & \centering{$2x+14$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}  \end{minipage}
\hfill
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$x^2-3$} & \centering{$2x+3$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular} \end{minipage}

\begin{table}[h!]
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|c}
\hline \centering{ $9-4x-9+7x$\\ \hspace{1cm}} & \centering{$37-13x$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}  \end{minipage}
\hfill
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$-3x+12$ \\ \hspace{1cm}} & \centering{$x(x-3)$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}
\end{minipage}\end{table}

\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$5$ \\ \hspace{1cm}} & \centering{$2x+7$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}  \end{minipage}
\hfill
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$2(x-4)+5(9-3x)$ \\ \hspace{1cm}} & \centering{$-6x^2+8x$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}  \end{minipage}

\begin{table}[h!]
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$-1+2(2+x)$ \\ \hspace{1cm}} & \centering{$3x-3$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}  \end{minipage}
\hfill
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}||p{3.5cm}|}
\hline \centering{$4(x-3)-4x+17$} & \centering{$3x$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular} \end{minipage}
\end{table}

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