Source
%@Auteur:Véronique Glaçon.\par
Après avoir découpés 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}

\renewcommand\arraystretch{2.5}

\begin{table}[h!]
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}|p{3.5cm}|}
\hline \centering{$x^2-10x+25$ \\ \hspace{1cm}} & \centering{$x(x+5)-4x$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}
\end{minipage}
\hfill
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}|p{3.5cm}|}
\hline \centering{$(1-2x)(2x+3)$ \\ \hspace{1cm}} & \centering{$15x-9$ \\ \hspace{1cm}} \tabularnewline \hline
\end{tabular}
\end{minipage}
\end{table}

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

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

%\begin{table}[h!]
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}|p{3.5cm}|}
\hline \centering{$3(5x-3)$ \\ \hspace{1cm}} & \centering{$(x-5)^2$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}
\end{minipage}
\hfill
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}|p{3.5cm}|}
\hline \centering{$4x(2x+7)$ \\ \hspace{1cm}} & \centering{$x^2+8x+15$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}
\end{minipage}
%\end{table}

\begin{table}[h!]
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}|p{3.5cm}|}
\hline \centering{$(5-3x)(6-2x)$ \\ \hspace{1cm}} & \centering{$8x^2+14x$ \\ \hspace{1cm}} \tabularnewline  \hline
\end{tabular}
\end{minipage}
\hfill
\begin{minipage}[t]{.5\linewidth}
\begin{tabular}{|p{3.5cm}|p{3.5cm}|}
\hline \centering{$6x^2-28x+30$} & \centering{$(1-2x)(5x+2)+(1-2x)(3x-5)$} \tabularnewline  \hline
\end{tabular}
\end{minipage}
\end{table}