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%%%
% Equations
%%%
\setKVdefault[ClesEquation]{Ecart=0.5,Fleches=false,FlecheDiv=false,Laurent=false,Decomposition=false,Terme=false,Composition=false,Symbole=false,Decimal=false,Entier=false,Lettre=x,Solution=false,LettreSol=true,Bloc=false,Simplification=false,CouleurTerme=black,CouleurCompo=black,CouleurSous=red,CouleurSymbole=orange,Verification=false,Nombre=0,Egalite=false,Produit=false,Facteurs=false,Carre=false,Exact=false,Pose=false,Equivalence=false}
\newcommand\rightcomment[4]{%
\begin{tikzpicture}[remember picture,overlay]
\draw[Cfleches,-stealth]
($({pic cs:#3}|-{pic cs:#1})+(\useKV[ClesEquation]{Ecart},0)$)
.. controls +(0.2,-0.05) and +(0.2,0.1) ..
node[right,align=left]{#4}
($({pic cs:#3}|-{pic cs:#2})+(\useKV[ClesEquation]{Ecart},0.1)$);
\end{tikzpicture}%
}%
\newcommand\leftcomment[4]{%
\begin{tikzpicture}[remember picture,overlay]
\draw[Cfleches,-stealth]
($({pic cs:#3}|-{pic cs:#1})-(\useKV[ClesEquation]{Ecart},0)$)
.. controls +(-0.2,-0.05) and +(-0.2,0.1) ..
node[left,align=right]{#4}
($({pic cs:#3}|-{pic cs:#2})-(\useKV[ClesEquation]{Ecart},-0.1)$);
\end{tikzpicture}%
}%
\newcommand\Rightcomment[4]{%
\begin{tikzpicture}[remember picture,overlay]
\draw[Cfleches,-stealth]
($({pic cs:#3}|-{pic cs:#1})+(\useKV[ClesEquation]{Ecart},0)$)
.. controls +(0.2,-0.05) and +(0.2,0.1) ..
node[right,align=left]{#4}
($({pic cs:#3}|-{pic cs:#2})+(\useKV[ClesEquation]{Ecart},0.1)$);
\end{tikzpicture}%
}%
\newcommand\Leftcomment[4]{%
\begin{tikzpicture}[remember picture,overlay]
\draw[Cfleches,-stealth]
($({pic cs:#3}|-{pic cs:#1})-(\useKV[ClesEquation]{Ecart},0)$)
.. controls +(-0.2,-0.05) and +(-0.2,0.1) ..
node[left,align=right]{#4}
($({pic cs:#3}|-{pic cs:#2})-(\useKV[ClesEquation]{Ecart},-0.1)$);
\end{tikzpicture}%
}%
% Pour "oublier" les tikzmarks. En cas de plusieurs utilisations de la macro \ResolEquation
\newcounter{Nbequa}
\setcounter{Nbequa}{0}
%CT
\newdimen\fdashwidth\fdashwidth = 0.8pt % \'epaisseur traits
\newdimen\fdashlength\fdashlength = 0.5mm % longueur des pointill\'es et s\'eparation entre pointill\'es
\newdimen\fdashsep\fdashsep = 3pt % s\'eparateur entre contenu et traits
\def\fdash#1{%
\leavevmode\begingroup%
\setbox0\hbox{#1}%
\def\hdash{\vrule height\fdashwidth width\fdashlength\relax}%
\def\vdash{\hrule height\fdashlength width\fdashwidth\relax}%
\def\dashblank{\kern\fdashlength}%
\ifdim\fdashsep>0pt%
\setbox0\hbox{\vrule width0pt height\dimexpr\ht0+\fdashsep depth\dimexpr\dp0+\fdashsep\kern\fdashsep\unhbox0 \kern\fdashsep}%
\fi%
\edef\hdash{\hbox to\the\wd0{\noexpand\color{Csymbole}\hdash\kern.5\fdashlength\xleaders\hbox{\hdash\dashblank}\hfil\hdash}}%
\edef\vdash{\vbox to\the\dimexpr\ht0+\dp0+2\fdashwidth{\noexpand\color{Csymbole}\vdash\kern.5\fdashlength\xleaders\vbox{\vdash\dashblank}\vfil\vdash}}%
\hbox{%
\vdash%
\vtop{\vbox{\offinterlineskip\hdash\hbox{\unhbox0 }\hdash}}%
\vdash}%
\endgroup%
}%
% fin CT
\def\Fdash#1{\raisebox{-2\fdashsep+\fdashwidth}{\fdash{#1}}}%
%Une simplification de a/b est possible ou non ?
\newboolean{Simplification}
\newcommand\SSimpliTest[2]{%
% Test d'une simplification possible ou pas de #1/#2
\newcount\numerateur\newcount\denominateur\newcount\valabsnum\newcount\valabsdeno%
\numerateur=\number#1
\denominateur=\number#2
\ifnum\number#1<0
\valabsnum=\numexpr0-\number#1
\else
\valabsnum=\number#1
\fi
\ifnum\number#2<0
\valabsdeno=\numexpr0-\number#2
\else
\valabsdeno=\number#2
\fi
\ifnum\the\valabsnum=0
\setboolean{Simplification}{true}
\else
\PGCD{\the\valabsnum}{\the\valabsdeno}
\ifnum\pgcd>1
\setboolean{Simplification}{true}
\else
\ifnum\the\numerateur<0
\ifnum\the\denominateur<0
\setboolean{Simplification}{true}
\else
\ifnum\valabsdeno=1\relax
\setboolean{Simplification}{true}
\else
\setboolean{Simplification}{false}
\fi
\fi
\else
\ifnum\valabsdeno=1\relax
\setboolean{Simplification}{true}
\else
\setboolean{Simplification}{false}
\fi
\fi
\fi
\fi
}
\definecolor{Cfleches}{RGB}{100,100,100}%
\newcommand\AffichageEqua[4]{%
\def\LETTRE{\useKV[ClesEquation]{Lettre}}%
\ensuremath{%
% partie du x
\xintifboolexpr{#1==0}{}{\xintifboolexpr{#1==1}{}{\xintifboolexpr{#1==-1}{-}{\num{#1}}}\LETTRE}%
% partie du nombre
\xintifboolexpr{#2==0}{\xintifboolexpr{#1==0}{0}{}}{\xintifboolexpr{#2>0}{\xintifboolexpr{#1==0}{}{+}\num{#2}}{\num{#2}}}%
% egal
=
% partie du x
\xintifboolexpr{#3==0}{}{\xintifboolexpr{#3==1}{}{\xintifboolexpr{#3==-1}{-}{\num{#3}}}\LETTRE}%
% partie du nombre
\xintifboolexpr{#4==0}{%
\xintifboolexpr{#3==0}{0}{}
}{%
\xintifboolexpr{#4>0}{\xintifboolexpr{#3==0}{}{+}\num{#4}}{\num{#4}}%
}
}%
}%
\newcommand\EcrireSolutionEquation[4]{%
L'équation \AffichageEqua{#1}{#2}{#3}{#4} a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.
}%
\input{PfCEquationSoustraction2}%
\input{PfCEquationTerme1}%
\input{PfCEquationComposition2}%
\input{PfCEquationPose1}%
\input{PfCEquationSymbole1}%
\input{PfCEquationLaurent1}
\newcommand\ResolEquation[5][]{%
\useKVdefault[ClesEquation]%
\setKV[ClesEquation]{#1}%
\colorlet{Cterme}{\useKV[ClesEquation]{CouleurTerme}}%
\colorlet{Ccompo}{\useKV[ClesEquation]{CouleurCompo}}%
\colorlet{Csymbole}{\useKV[ClesEquation]{CouleurSymbole}}%
\colorlet{Cdecomp}{\useKV[ClesEquation]{CouleurSous}}%
\ifboolKV[ClesEquation]{Carre}{%
\ResolEquationCarre[#1]{#2}%
}{%
\ifboolKV[ClesEquation]{Produit}{%
\ResolEquationProduit[#1]{#2}{#3}{#4}{#5}%
}{%
\ifboolKV[ClesEquation]{Verification}{%
\Verification[#1]{#2}{#3}{#4}{#5}%
}{%
\ifboolKV[ClesEquation]{Symbole}{%
\ResolEquationSymbole[#1]{#2}{#3}{#4}{#5}%
}{%
\ifboolKV[ClesEquation]{Laurent}{%
\ResolEquationLaurent[#1]{#2}{#3}{#4}{#5}%
}{%
\ifboolKV[ClesEquation]{Terme}{%
\ResolEquationTerme[#1]{#2}{#3}{#4}{#5}%
}{\ifboolKV[ClesEquation]{Composition}{%
\ResolEquationComposition[#1]{#2}{#3}{#4}{#5}%
}{\ifboolKV[ClesEquation]{Pose}{%
\ResolEquationL[#1]{#2}{#3}{#4}{#5}%
}{%
\ResolEquationSoustraction[#1]{#2}{#3}{#4}{#5}%
}%
}%
}%
}%
}%
}%
}%
}%
}%
\newcommand\ResolEquationCarre[2][]{%
\setKV[ClesEquation]{#1}%
\xintifboolexpr{#2<0}{%
Comme $\num{#2}$ est n\'egatif, alors l'\'equation $\useKV[ClesEquation]{Lettre}^2=\num{#2}$ n'a aucune solution.%
}{\xintifboolexpr{#2==0}{%
L'\'equation $\useKV[ClesEquation]{Lettre}^2=0$ a une unique solution : $\useKV[ClesEquation]{Lettre}=0$.%
}{%
Comme \num{#2} est positif, alors l'\'equation $\useKV[ClesEquation]{Lettre}^2=\num{#2}$ a deux solutions :%
\begin{align*}
\useKV[ClesEquation]{Lettre}&=\sqrt{\num{#2}}&&\text{et}&\useKV[ClesEquation]{Lettre}&=-\sqrt{\num{#2}}%\\
\ifboolKV[ClesEquation]{Exact}{\\%
\useKV[ClesEquation]{Lettre}&=\num{\fpeval{sqrt(#2)}}&&\text{et}&\useKV[ClesEquation]{Lettre}&=-\num{\fpeval{sqrt(#2)}}}{}%
\end{align*}
}%
}%
}%
\newcommand\ResolEquationProduit[5][]{%
\setKV[ClesEquation]{#1}%
\ifboolKV[ClesEquation]{Equivalence}{}{C'est un produit nul donc \ifboolKV[ClesEquation]{Facteurs}{l'un au
moins des facteurs est nul}{} :}%
\ifboolKV[ClesEquation]{Equivalence}{%
\[\Distri{#2}{#3}{#4}{#5}=0\]
\begin{align*}%
&\makebox[0pt]{$\Longleftrightarrow$}&\xintifboolexpr{#3==0}{\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}}{\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}}&=0&\quad&\makebox[0pt]{ou}\quad&\xintifboolexpr{#5==0}{\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}}{\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}}&=0\\
&\makebox[0pt]{$\Longleftrightarrow$}&\xintifboolexpr{#3==0}{\xdef\Coeffa{1}\xdef\Coeffb{\fpeval{0-#3}}\xintifboolexpr{#2==1}{&}{\useKV[ClesEquation]{Lettre}&=0}}{\xdef\Coeffa{#2}\xdef\Coeffb{\fpeval{0-#3}}\xintifboolexpr{\Coeffa==1}{}{\num{\Coeffa}}\useKV[ClesEquation]{Lettre}&=\num{\Coeffb}}&&&\xintifboolexpr{#5==0}{\xdef\Coeffc{1}\xdef\Coeffd{\fpeval{0-#5}}\xintifboolexpr{#4==1}{&}{\useKV[ClesEquation]{Lettre}&=0}}{\xdef\Coeffc{#4}\xdef\Coeffd{\fpeval{0-#5}}\xintifboolexpr{\Coeffc==1}{}{\num{\Coeffc}}\useKV[ClesEquation]{Lettre}&=\num{\Coeffd}}%\\
\xintifboolexpr{\Coeffa==1 'and' \Coeffc==1}{}{\\%\ifnum\cmtd>1
&\makebox[0pt]{$\Longleftrightarrow$}&\xintifboolexpr{\Coeffa==1}{&}{\useKV[ClesEquation]{Lettre}&=\frac{\num{\Coeffb}}{\num{\Coeffa}}}\xintifboolexpr{\Coeffc==1}{}{&&&\useKV[ClesEquation]{Lettre}&=\frac{\num{\Coeffd}}{\num{\Coeffc}}}
% accolade%\\
%%%%
\ifboolKV[ClesEquation]{Entier}{%
\xdef\TSimp{}%
\SSimpliTest{\Coeffb}{\Coeffa}\ifthenelse{\boolean{Simplification}}{\xintifboolexpr{#3==0}{\xdef\TSimp{0}}{\xdef\TSimp{1}}}{\xdef\TSimp{0}}
\SSimpliTest{\Coeffd}{\Coeffc}\ifthenelse{\boolean{Simplification}}{\xintifboolexpr{#5==0}{}{\xdef\TSimp{\fpeval{\TSimp+1}}}}{}
\xintifboolexpr{\TSimp==0}{}{\\
\ifboolKV[ClesEquation]{Simplification}{%
&\makebox[0pt]{$\Longleftrightarrow$}&\SSimpliTest{\Coeffb}{\Coeffa}\xintifboolexpr{\Coeffa==1}{&}{\ifthenelse{\boolean{Simplification}}{\useKV[ClesEquation]{Lettre}&=\SSimplifie{\Coeffb}{\Coeffa}}{&}%\\
}
}{}
&&&\ifboolKV[ClesEquation]{Simplification}{%
\SSimpliTest{\Coeffd}{\Coeffc}%
\xintifboolexpr{\Coeffc==1}{}{\ifthenelse{\boolean{Simplification}}{\useKV[ClesEquation]{Lettre}&=\SSimplifie{\Coeffd}{\Coeffc}}{}%\\
}
}{}
}
}{}
}
\end{align*}
}{%
\begin{align*}
\xintifboolexpr{#3==0}{\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}}{\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}}&=0&&\text{ou}&\xintifboolexpr{#5==0}{\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}}{\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}}&=0\\
\xintifboolexpr{#3==0}{\xdef\Coeffa{1}\xdef\Coeffb{\fpeval{0-#3}}\xintifboolexpr{#2==1}{&}{\useKV[ClesEquation]{Lettre}&=0}}{\xdef\Coeffa{#2}\xdef\Coeffb{\fpeval{0-#3}}\xintifboolexpr{\Coeffa==1}{}{\num{\Coeffa}}\useKV[ClesEquation]{Lettre}&=\num{\Coeffb}}&&&\xintifboolexpr{#5==0}{\xdef\Coeffc{1}\xdef\Coeffd{\fpeval{0-#5}}\xintifboolexpr{#4==1}{&}{\useKV[ClesEquation]{Lettre}&=0}}{\xdef\Coeffc{#4}\xdef\Coeffd{\fpeval{0-#5}}\xintifboolexpr{\Coeffc==1}{}{\num{\Coeffc}}\useKV[ClesEquation]{Lettre}&=\num{\Coeffd}}%\\
\xintifboolexpr{\Coeffa==1 'and' \Coeffc==1}{}{\\%\ifnum\cmtd>1
\xintifboolexpr{\Coeffa==1}{&}{\useKV[ClesEquation]{Lettre}&=\frac{\num{\Coeffb}}{\num{\Coeffa}}}\xintifboolexpr{\Coeffc==1}{}{&&&\useKV[ClesEquation]{Lettre}&=\frac{\num{\Coeffd}}{\num{\Coeffc}}}
%accolade%\\
%%%%
\ifboolKV[ClesEquation]{Entier}{%
\xdef\TSimp{}
\SSimpliTest{\Coeffb}{\Coeffa}\ifthenelse{\boolean{Simplification}}{\xintifboolexpr{#3==0}{\xdef\TSimp{0}}{\xdef\TSimp{1}}}{\xdef\TSimp{0}}
\SSimpliTest{\Coeffd}{\Coeffc}\ifthenelse{\boolean{Simplification}}{\xintifboolexpr{#5==0}{}{\xdef\TSimp{\fpeval{\TSimp+1}}}}{}
\xintifboolexpr{\TSimp==0}{}{\\
\ifboolKV[ClesEquation]{Simplification}{%
\SSimpliTest{\Coeffb}{\Coeffa}
\xintifboolexpr{\Coeffa==1}{&}{\ifthenelse{\boolean{Simplification}}{\useKV[ClesEquation]{Lettre}&=\SSimplifie{\Coeffb}{\Coeffa}}{&}%\\
}
}{}
&&&\ifboolKV[ClesEquation]{Simplification}{%
\SSimpliTest{\Coeffd}{\Coeffc}%
\xintifboolexpr{\Coeffc==1}{}{\ifthenelse{\boolean{Simplification}}{\useKV[ClesEquation]{Lettre}&=\SSimplifie{\Coeffd}{\Coeffc}}{}%\\
}
}{}
}
}{}
}
\end{align*}
}%
\ifboolKV[ClesEquation]{Solution}{L'\'equation $\xintifboolexpr{#3==0}{\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}}{(\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}})}\xintifboolexpr{#5==0}{\times\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}}{(\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}})}=0$ a deux solutions : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$ et \opdiv*{\Coeffd}{\Coeffc}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffd}{\Coeffc}}{\frac{\num{\Coeffd}}{\num{\Coeffc}}}\fi$.
}{}%
}
\newcommand\Verification[5][]{%
\setKV[ClesEquation]{#1}%
\xdef\ValeurTest{\useKV[ClesEquation]{Nombre}}%
Testons la valeur $\useKV[ClesEquation]{Lettre}=\num{\ValeurTest}$ :%
\begin{align*}
\xintifboolexpr{#2==0}{\num{#3}}{\num{#2}\times\xintifboolexpr{\ValeurTest<0}{(\num{\ValeurTest})}{\num{\ValeurTest}}\xintifboolexpr{#3==0}{}{\xintifboolexpr{#3>0}{+\num{#3}}{\num{#3}}}}&&\xintifboolexpr{#4==0}{\num{#5}}{\num{#4}\times\xintifboolexpr{\ValeurTest<0}{(\num{\ValeurTest})}{\num{\ValeurTest}}\xintifboolexpr{#5==0}{}{\xintifboolexpr{#5>0}{+\num{#5}}{\num{#5}}}}\\
\xintifboolexpr{#2==0}{}{\num{\fpeval{#2*\useKV[ClesEquation]{Nombre}}}\xintifboolexpr{#3==0}{}{\xintifboolexpr{#3>0}{+\num{#3}}{\num{#3}}}}&&\xintifboolexpr{#4==0}{}{\num{\fpeval{#4*\useKV[ClesEquation]{Nombre}}}\xintifboolexpr{#5==0}{}{\xintifboolexpr{#5>0}{+\num{#5}}{\num{#5}}}}\\
\xintifboolexpr{#2==0}{}{\num{\fpeval{#2*\useKV[ClesEquation]{Nombre}+#3}}}&&\xintifboolexpr{#4==0}{}{\num{\fpeval{#4*\useKV[ClesEquation]{Nombre}+#5}}}
\end{align*}
\xdef\Testa{\fpeval{#2*\useKV[ClesEquation]{Nombre}+#3}}\xdef\Testb{\fpeval{#4*\useKV[ClesEquation]{Nombre}+#5}}%
\ifboolKV[ClesEquation]{Egalite}{%
Comme \xintifboolexpr{\Testa==\Testb}{$\num{\Testa}=\num{\Testb}$}{$\num{\Testa}\not=\num{\Testb}$}, alors l'\'egalit\'e $\xintifboolexpr{#2==0}{\num{#3}}{\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3==0}{}{\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}}}=\xintifboolexpr{#4==0}{\num{#5}}{\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5==0}{}{\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}}}$ \xintifboolexpr{\Testa==\Testb}{ est v\'erifi\'ee }{ n'est pas v\'erifi\'ee } pour $\useKV[ClesEquation]{Lettre}=\num{\useKV[ClesEquation]{Nombre}}$.%
}{\xintifboolexpr{\Testa==\Testb}{Comme $\num{\Testa}=\num{\Testb}$, alors $\useKV[ClesEquation]{Lettre}=\num{\useKV[ClesEquation]{Nombre}}$ est bien }{Comme $\num{\Testa}\not=\num{\Testb}$, alors $\useKV[ClesEquation]{Lettre}=\num{\useKV[ClesEquation]{Nombre}}$ n'est pas }une solution de l'\'equation $\xintifboolexpr{#2==0}{\num{#3}}{\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3==0}{}{\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}}}=\xintifboolexpr{#4==0}{\num{#5}}{\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5==0}{}{\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}}}$.}%
}%
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