Refonte de la documentation. Ajout d'une option de coloriage en z dans les surfaces...

[mp-solid.git] / doc / Christ5.tex

%=========================================
%Macros personnelles
%christophe.poulain@melusine.eu.org
%création : 25 Septembre 1999
%dernière modification : 28 Avril 2005
%=========================================

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\input{xlopsqrt}
\newtheorem{ppte}{Propri\'et\'e}
\newtheorem{theo}{Th\'eor\`eme}
\newtheorem{defi}{D\'efinition}
\newtheorem{lemme}{Lemme}
\newtheorem{coro}{Corollaire}
\newtheorem{prop}{Proposition}
\newtheorem{reg}{R\`egle}
\newtheorem{conj}{Conjecture}
\newtheorem{remar}{Remarque}
\newtheorem{exem}{Exemple}

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\par\underline{\bf Exercice~\thenum} }

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\newboolean{exact}
\setboolean{exact}{true}
\newboolean{racine}
\setboolean{racine}{false}

\opcopy{2}{pres}

\newcommand{\pythahypo}[5]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#4}{A1}%
\opcopy{#5}{A2}%
Dans le triangle $#1#2#3$ rectangle en $#2$, le th\'eor\`eme de Pythagore permet d'\'ecrire :%
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
#1#3^2&=\opprint{A1}^2+\opprint{A2}^2\cr
#1#3^2&=\opmul*{A1}{A1}{a1}\opprint{a1}+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#3^2&=\opadd*{a1}{a2}{a3}\opprint{a3}\cr
#1#3&=\sqrt{\opprint{a3}}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#3&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#3&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\egapythahypo}[5]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#4}{A1}%
\opcopy{#5}{A2}%
Comme le triangle $#1#2#3$ est rectangle en $#2$, alors l'égalité de
Pythagore est vérifiée :
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
#1#3^2&=\opprint{A1}^2+\opprint{A2}^2\cr
#1#3^2&=\opmul*{A1}{A1}{a1}\opprint{a1}+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#3^2&=\opadd*{a1}{a2}{a3}\opprint{a3}\cr
#1#3&=\sqrt{\opprint{a3}}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#3&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#3&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\egapythahyposansrac}[5]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#4}{A1}%
\opcopy{#5}{A2}%
Comme le triangle $#1#2#3$ est rectangle en $#2$, alors l'égalité de
Pythagore est vérifiée :
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
#1#3^2&=\opprint{A1}^2+\opprint{A2}^2\cr
#1#3^2&=\opmul*{A1}{A1}{a1}\opprint{a1}+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#3^2&=\opadd*{a1}{a2}{a3}\opprint{a3}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#3&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#3&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\egapythahyposansracd}[5]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#4}{A1}%
\opcopy{#5}{A2}%
Le triangle $#1#2#3$ est rectangle en $#2$ donc l'égalité de
Pythagore est vérifiée :
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
#1#3^2&=\opprint{A1}^2+\opprint{A2}^2\cr
#1#3^2&=\opmul*{A1}{A1}{a1}\opprint{a1}+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#3^2&=\opadd*{a1}{a2}{a3}\opprint{a3}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#3&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#3&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\egapythahypod}[5]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#4}{A1}%
\opcopy{#5}{A2}%
Le triangle $#1#2#3$ est rectangle en $#2$ donc l'égalité de
Pythagore est vérifiée :
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
#1#3^2&=\opprint{A1}^2+\opprint{A2}^2\cr
#1#3^2&=\opmul*{A1}{A1}{a1}\opprint{a1}+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#3^2&=\opadd*{a1}{a2}{a3}\opprint{a3}\cr
#1#3&=\sqrt{\opprint{a3}}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#3&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#3&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\pythadroit}[5]{
\opset{decimalsepsymbol={,}}
\opcopy{#4}{A1}
\opcopy{#5}{A2}
Dans le triangle $#1#2#3$ rectangle en $#2$, le théorème de Pythagore permet d'écrire :
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
\opprint{A1}^2&=#1#2^2+\opprint{A2}^2\cr
\opmul*{A1}{A1}{a1}\opprint{a1}&=#1#2^2+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#2^2&=\opmul*{A1}{A1}{a1}\opprint{a1}-\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#2^2&=\opsub*{a1}{a2}{a3}\opprint{a3}\cr
#1#2&=\sqrt{\opprint{a3}}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#2&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#2&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\egapythadroit}[5]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#4}{A1}%
\opcopy{#5}{A2}%
Comme le triangle $#1#2#3$ est rectangle en $#2$, alors l'égalité de %
Pythagore est vérifiée :%
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
\opprint{A1}^2&=#1#2^2+\opprint{A2}^2\cr
\opmul*{A1}{A1}{a1}\opprint{a1}&=#1#2^2+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#2^2&=\opmul*{A1}{A1}{a1}\opprint{a1}-\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#2^2&=\opsub*{a1}{a2}{a3}\opprint{a3}\cr
#1#2&=\sqrt{\opprint{a3}}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#2&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#2&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\egapythadroitsansrac}[5]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#4}{A1}%
\opcopy{#5}{A2}%
Comme le triangle $#1#2#3$ est rectangle en $#2$ alors l'égalité de %
Pythagore est vérifiée :%
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
\opprint{A1}^2&=#1#2^2+\opprint{A2}^2\cr
\opmul*{A1}{A1}{a1}\opprint{a1}&=#1#2^2+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#2^2&=\opmul*{A1}{A1}{a1}\opprint{a1}-\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#2^2&=\opsub*{a1}{a2}{a3}\opprint{a3}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#2&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#2&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\egapythadroitd}[5]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#4}{A1}%
\opcopy{#5}{A2}%
Le triangle $#1#2#3$ est rectangle en $#2$ donc l'égalité de %
Pythagore est vérifiée :
\[\Eqalign{
#1#3^2&=#1#2^2+#2#3^2\cr
\opprint{A1}^2&=#1#2^2+\opprint{A2}^2\cr
\opmul*{A1}{A1}{a1}\opprint{a1}&=#1#2^2+\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#2^2&=\opmul*{A1}{A1}{a1}\opprint{a1}-\opmul*{A2}{A2}{a2}\opprint{a2}\cr
#1#2^2&=\opsub*{a1}{a2}{a3}\opprint{a3}\cr
#1#2&=\sqrt{\opprint{a3}}\cr
\ifthenelse{\boolean{racine}}{}{\ifthenelse{\boolean{exact}}{#1#2&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}{#1#2&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{pres}{a4}\opunzero{a4}
\opprint{a4}}}\cr
}\]
}

\newcommand{\Recipytha}[6]{
\opset{decimalsepsymbol={,}}
\opcopy{#4}{A1}
\opcopy{#5}{A2}
\opcopy{#6}{A3}
Dans le triangle $#1#2#3$, $[#1#3]$ est le plus grand côté.
\[\left.
  \begin{array}{l}
    #1#3^2=\opprint{A1}^2=\opmul*{A1}{A1}{a1}\opprint{a1}\cr
    #1#2^2+#2#3^2=\opprint{A2}^2+\opprint{A3}^2=\opmul*{A2}{A2}{a2}\opprint{a2}+\opmul*{A3}{A3}{a3}\opprint{a3}=\opadd*{a2}{a3}{a4}\opprint{a4}\cr
  \end{array}
\right\}#1#3^2=#1#2^2+#2#3^2
\]
Comme $#1#3^2=#1#2^2+#2#3^2$ alors le triangle $#1#2#3$ est rectangle
en $#2$ d'après la réciproque du théorème de Pythagore.}

\newcommand{\Reciegapytha}[6]{
\opset{decimalsepsymbol={,}}
\opcopy{#4}{A1}
\opcopy{#5}{A2}
\opcopy{#6}{A3}
Dans le triangle $#1#2#3$, $[#1#3]$ est le plus grand côté.
\[\left.
  \begin{array}{l}
    #1#3^2=\opprint{A1}^2=\opmul*{A1}{A1}{a1}\opprint{a1}\cr
    #1#2^2+#2#3^2=\opprint{A2}^2+\opprint{A3}^2=\opmul*{A2}{A2}{a2}\opprint{a2}+\opmul*{A3}{A3}{a3}\opprint{a3}=\opadd*{a2}{a3}{a4}\opprint{a4}\cr
  \end{array}
\right\}#1#3^2=#1#2^2+#2#3^2
\]
L'égalité de Pythagore est vérifiée donc le triangle $#1#2#3$ est rectangle en $#2$.}

\newcommand{\Recipythacol}[6]{
\opset{decimalsepsymbol={,}}
\opcopy{#4}{A1}
\opcopy{#5}{A2}
\opcopy{#6}{A3}
Dans le triangle $#1#2#3$, $[#1#3]$ est le plus grand côté.
\[\Eqalign{
    #1#3^2&\kern0.15\linewidth&#1#2^2+#2#3^2\cr
    \opprint{A1}^2&&\opprint{A2}^2+\opprint{A3}^2\cr
    \opmul*{A1}{A1}{a1}\opprint{a1}&&\opmul*{A2}{A2}{a2}\opprint{a2}+\opmul*{A3}{A3}{a3}\opprint{a3}\cr
    &&\opadd*{a2}{a3}{a4}\opprint{a4}\cr
}\]
Comme $#1#3^2=#1#2^2+#2#3^2$ alors le triangle $#1#2#3$ est rectangle
en $#2$ d'après la réciproque du théorème de Pythagore.}

\newcommand{\Reciegapythacol}[6]{
\opset{decimalsepsymbol={,}}
\opcopy{#4}{A1}
\opcopy{#5}{A2}
\opcopy{#6}{A3}
Dans le triangle $#1#2#3$, $[#1#3]$ est le plus grand côté.
\[\Eqalign{
    #1#3^2&\kern0.15\linewidth&#1#2^2+#2#3^2\cr
    \opprint{A1}^2&&\opprint{A2}^2+\opprint{A3}^2\cr
    \opmul*{A1}{A1}{a1}\opprint{a1}&&\opmul*{A2}{A2}{a2}\opprint{a2}+\opmul*{A3}{A3}{a3}\opprint{a3}\cr
    &&\opadd*{a2}{a3}{a4}\opprint{a4}\cr
}\]
Comme l'égalité de Pythagore est vérifiée alors le triangle $#1#2#3$ est rectangle en $#2$.}

\newcommand{\Thales}[5]{%
Dans le triangle $#1#2#3$, $#4$ est un point de la
 droite $(#1#2)$, $#5$ est un point de la droite
 $(#1#3)$ ; les droites $(#4#5)$ et $(#2#3)$ sont parallèles.
 Le théorème de Thalès permet d'écrire :
\[\frac{#1#4}{#1#2}=\frac{#1#5}{#1#3}=\frac{#4#5}{#2#3}\]%
}

\newcommand{\Thalesd}[5]{%
Dans le triangle $#1#2#3$, $#4$ appartient à la
 droite $(#1#2)$, $#5$ appartient à la droite
 $(#1#3)$. Comme les droites $(#4#5)$ et $(#2#3)$ sont parallèles
 alors le théorème de Thalès permet d'écrire :
\[\frac{#1#4}{#1#2}=\frac{#1#5}{#1#3}=\frac{#4#5}{#2#3}\]%
}

\newcommand{\Thalesa}[5]{%
Dans le triangle $#1#2#3$, $#4$ appartient à la
 droite $(#1#2)$, $#5$ appartient à la droite
 $(#1#3)$ ; les droites $(#4#5)$ et $(#2#3)$ sont parallèles.
 Le théorème de Thalès permet d'écrire :
\[\frac{#1#4}{#1#2}=\frac{#1#5}{#1#3}=\frac{#4#5}{#2#3}\]%
}

\newcommand{\Thalesf}[5]{
Dans le triangle $#1#2#3$, $#4$ est un point du
 segment $[#1#2]$, $#5$ est un point du segment
 $[#1#3]$ ; les droites $(#4#5)$ et $(#2#3)$ sont parallèles.
 L'égalité des 3 rapports permet d'écrire :
\[\frac{#1#4}{#1#2}=\frac{#1#5}{#1#3}=\frac{#4#5}{#2#3}\]
}

\newcommand{\ThalesF}[5]{
Dans le triangle $#1#2#3$, $#4$ est un point du
 segment $[#1#2]$, $#5$ est un point du segment
 $[#1#3]$. Comme les droites $(#4#5)$ et $(#2#3)$ sont parallèles
 alors, d'après l'égalité des trois rapports, on a :
\[\frac{#1#4}{#1#2}=\frac{#1#5}{#1#3}=\frac{#4#5}{#2#3}.\]
}

\newcommand{\ResolThales}[6]{%
\opset{decimalsepsymbol={,}}%
\opcopy{#3}{a3}%
\opcopy{#4}{a4}%
\opcopy{#5}{a5}%
On utilise %
\[\Eqalign{%
\frac{#1#2}{\opprint{a3}}&=\frac{\opprint{a4}}{\opprint{a5}}\cr%
#1#2&=\frac{\opprint{a3}\times\opprint{a4}}{\opprint{a5}}\cr%
#1#2&=\frac{\opmul*{a3}{a4}{a6}\opunzero{a6}\opprint{a6}}{\opprint{a5}}\cr%
\ifthenelse{\boolean{exact}}{#1#2&=\opdiv*[maxdivstep=4]{a6}{a5}{a7}{a8}\opunzero{a7}\opprint{a7}\cr}{#1#2&\approx\opdiv*[maxdivstep=4]{a6}{a5}{a7}{a8}\opunzero{a7}\opprint{a7}\cr}%
}\]%
\ifthenelse{\boolean{exact}}{La longueur $#1#2$ mesure \opprint{a7}\,#6}{La longueur $#1#2$ mesure environ \opprint{a7}\,#6}%
}

%%QCM
\newcounter{qqcm}

%définir un booléen qui permet de choisir la correction ou non
\newboolean{solution}

%définir une commande \V qui permet de changer le carré en carré coché suivant la valeur du booléen.
\newcommand{\V}[1]{\ifthenelse{\boolean{solution}}{$\boxtimes$\kern2mm #1}{$\Box$\kern2mm #1}}
\newcommand{\F}[1]{$\Box$\kern2mm #1}
\newcommand{\vr}{\ifthenelse{\boolean{solution}}{$\boxtimes$}{$\Box$}}
\newcommand{\fa}{$\Box$}

%%QCM Version 2
\newenvironment{Qcm}[1][2]{\par\setboolean{solution}{false}
\setcounter{qqcm}{0}\renewcommand{\arraystretch}{1.5}
\begin{tabular}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}b{\linewidth/#1}|*{#1}{l|}}\hline}{\hline\end{tabular}
\renewcommand{\arraystretch}{1}}

\newenvironment{Qcmcor}[1][2]{\par\setboolean{solution}{true}\setcounter{qqcm}{0}\renewcommand{\arraystretch}{1.5}
\begin{tabular}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}b{\linewidth/#1}|*{#1}{l|}}\hline}{\hline\end{tabular}
\renewcommand{\arraystretch}{1}}

\newcounter{taill}
\newcommand{\QCM}[3]{\setboolean{solution}{false}
\setcounter{qqcm}{0}
\renewcommand{\arraystretch}{1.5}
\setcounter{taill}{#1}
\addtocounter{taill}{1}
\begin{tabularx}{\linewidth}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}X|*{#1}{l|}}
\hline
\multicolumn{\thetaill}{|c|}{{\sc #2}}\\
#3
\hline
\end{tabularx}
\renewcommand{\arraystretch}{1}
}

\newcommand{\QCMcor}[3]{\setboolean{solution}{true}
\setcounter{qqcm}{0}
\renewcommand{\arraystretch}{1.5}
\setcounter{taill}{#1}
\addtocounter{taill}{1}
\begin{tabularx}{\linewidth}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}X|*{#1}{l|}}
\hline
\multicolumn{\thetaill}{|c|}{{\sc #2}}\\
#3
\hline
\end{tabularx}
\renewcommand{\arraystretch}{1}
}

\newcommand{\QCMvar}[4]{\setboolean{solution}{false}
\setcounter{qqcm}{0}
\renewcommand{\arraystretch}{#2}
\setcounter{taill}{#1}
\addtocounter{taill}{1}
\begin{tabularx}{\linewidth}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}X|*{#1}{l|}}
\hline
\multicolumn{\thetaill}{|c|}{{\sc #3}}\\
#4
\hline
\end{tabularx}
\renewcommand{\arraystretch}{1}%
}

\newcommand{\QCMvarcor}[4]{\setboolean{solution}{true}
\setcounter{qqcm}{0}
\renewcommand{\arraystretch}{#2}
\setcounter{taill}{#1}
\addtocounter{taill}{1}
\begin{tabularx}{\linewidth}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}X|*{#1}{l|}}
\hline
\multicolumn{\thetaill}{|c|}{{\sc #3}}\\
#4
\hline
\end{tabularx}
\renewcommand{\arraystretch}{1}
}

\newenvironment{VF}[1]{\par\setboolean{solution}{false}
\setcounter{qqcm}{0}\renewcommand{\arraystretch}{1.5}
\begin{center}
\begin{tabular}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}b{\linewidth/2}|*{2}{c|}}\hline
\multicolumn{3}{|c|}{#1}\\
\hline
\multicolumn{1}{|c|}{\bf Question}&\multicolumn{1}{c|}{\bf Vrai}&\multicolumn{1}{c|}{\bf Faux}\\
\hline
}{\hline\end{tabular}
\end{center}\renewcommand{\arraystretch}{1}}

\newenvironment{VFvar}[2]{\par\setboolean{solution}{false}
\setcounter{qqcm}{0}\renewcommand{\arraystretch}{1.5}
\begin{center}
\begin{tabular}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}b{#2}|*{2}{c|}}\hline
\multicolumn{3}{|c|}{#1}\\
\hline
\multicolumn{1}{|c|}{\bf Question}&\multicolumn{1}{c|}{\bf Vrai}&\multicolumn{1}{c|}{\bf Faux}\\
\hline
}{\hline\end{tabular}
\end{center}\renewcommand{\arraystretch}{1}}


\newenvironment{VFcor}[1]{\par\setboolean{solution}{true}\setcounter{qqcm}{0}\renewcommand{\arraystretch}{1.5}
\begin{tabular}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}b{\linewidth/2}|*{2}{c|}}\hline
\multicolumn{3}{|c|}{#1}\\
\hline
\multicolumn{1}{|c|}{\bf Question}&\multicolumn{1}{c|}{\bf Vrai}&\multicolumn{1}{c|}{\bf Faux}\\
\hline
}{\hline\end{tabular}
\renewcommand{\arraystretch}{1}}


\newcommand{\QCMsimple}[2]{\setboolean{solution}{false}
\setcounter{qqcm}{0}
\renewcommand{\arraystretch}{1.5}
\setcounter{taill}{#1}
\addtocounter{taill}{1}
\begin{tabularx}{\linewidth}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}X|*{#1}{l|}}
\hline
#2
\hline
\end{tabularx}
\renewcommand{\arraystretch}{1}
}

\newcommand{\QCMsimplevar}[3]{\setboolean{solution}{false}
\setcounter{qqcm}{0}
\renewcommand{\arraystretch}{#2}
\setcounter{taill}{#1}
\addtocounter{taill}{1}
\begin{tabularx}{\linewidth}{|>{\small\stepcounter{qqcm}{\bf \theqqcm$\blacktriangleright$}\,}X|*{#1}{l|}}
\hline
#3
\hline
\end{tabularx}
\renewcommand{\arraystretch}{1}
}

\newcommand{\QCMsimplecor}[2]{\setboolean{solution}{true}
\setcounter{qqcm}{0}
\renewcommand{\arraystretch}{1.5}
\setcounter{taill}{#1}
\addtocounter{taill}{1}
\begin{tabularx}{\linewidth}{|>{\small\stepcounter{qqcm}{\bf \theqqcm$\blacktriangleright$}\,}X|*{#1}{l|}}
\hline
#2
\hline
\end{tabularx}
\renewcommand{\arraystretch}{1}
}

\newcommand{\QCMsimplevarcor}[3]{\setboolean{solution}{true}
\setcounter{qqcm}{0}
\renewcommand{\arraystretch}{#2}
\setcounter{taill}{#1}
\addtocounter{taill}{1}
\begin{tabularx}{\linewidth}{|>{\small\stepcounter{qqcm}{\bf \theqqcm/}\,}X|*{#1}{l|}}
\hline
#3
\hline
\end{tabularx}
\renewcommand{\arraystretch}{1}
}