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-rw-r--r--macros/latex/contrib/profcollege/latex/PfCEquationPose1.tex49
1 files changed, 24 insertions, 25 deletions
diff --git a/macros/latex/contrib/profcollege/latex/PfCEquationPose1.tex b/macros/latex/contrib/profcollege/latex/PfCEquationPose1.tex
index 8bc7bb339f..735c09839d 100644
--- a/macros/latex/contrib/profcollege/latex/PfCEquationPose1.tex
+++ b/macros/latex/contrib/profcollege/latex/PfCEquationPose1.tex
@@ -1,12 +1,12 @@
% Licence : Released under the LaTeX Project Public License v1.3c
% or later, see http://www.latex-project.org/lppl.txtf
-\newcommand{\EquaBaseL}[5][]{%type ax=d ou b=cx
+\newcommand\EquaBaseL[5][]{%type ax=d ou b=cx
\useKVdefault[ClesEquation]%
\setKV[ClesEquation]{#1}%
\ifx\bla#2\bla%on teste si le paramètre #2 est vide:
% si oui, on est dans le cas b=cx. Eh bien on échange :)
% Mais attention si les deux paramètres a et c sont vides...
- \EquaBaseL[#1]{#4}{}{}{#3}
+ \EquaBaseL[#1]{#4}{0}{0}{#3}
\else
% si non, on est dans le cas ax=d
\xintifboolexpr{#2==0}{%
@@ -30,14 +30,15 @@
}{}
}{}
\end{align*}
- \ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{\useKV[ClesEquation]{Lettre}=\num{#5}}{\num{#2}\useKV[ClesEquation]{Lettre}=\num{#5}}$ a une unique solution : $\displaystyle\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\opdiv*{#5}{#2}{numequa}{resteequa}\opcmp{resteequa}{0}\ifopeq\opexport{numequa}{\numequa}\num{\numequa}\else\ifboolKV[ClesEquation]{Simplification}{\SSimplifie{#5}{#2}}{\frac{\num{#5}}{\num{#2}}}\fi$.%
- }{}
+ \xdef\Coeffa{#2}%
+ \xdef\Coeffb{#5}%
+ \ifboolKV[ClesEquation]{Solution}{\EcrireSolutionEquation{#2}{#3}{#4}{#5}}{}%
}
}
\fi
}
-\newcommand{\EquaDeuxL}[5][]{%type ax+b=d ou b=cx+d$
+\newcommand\EquaDeuxL[5][]{%type ax+b=d ou b=cx+d$
\useKVdefault[ClesEquation]%
\setKV[ClesEquation]{#1}%
\ifx\bla#2\bla%On échange en faisant attention à ne pas boucler : c doit être non vide
@@ -51,7 +52,7 @@
}%
}{%ELSE
\xintifboolexpr{#3==0}{%ax+b=d
- \EquaBaseL[#1]{#2}{}{}{#5}%
+ \EquaBaseL[#1]{#2}{0}{0}{#5}%
}{%ax+b=d$ Ici
\begin{align*}
\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}&=\num{#5}\\
@@ -76,28 +77,27 @@
}{}
}{}
\end{align*}
- \ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\num{#5}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.
- }{}
+ \ifboolKV[ClesEquation]{Solution}{\EcrireSolutionEquation{#2}{#3}{#4}{#5}}{}%
}
}
\fi
}
-\newcommand{\EquaTroisL}[5][]{%ax+b=cx ou ax=cx+d
+\newcommand\EquaTroisL[5][]{%ax+b=cx ou ax=cx+d
\useKVdefault[ClesEquation]%
\setKV[ClesEquation]{#1}%
\ifx\bla#3\bla%on inverse en faisant attention à la boucle #3<->#5
\ifx\bla#5\bla%
%% paramètre oublié
\else
- \EquaTroisL[#1]{#4}{#5}{#2}{}%
+ \EquaTroisL[#1]{#4}{#5}{#2}{0}%
\fi
\else
\xintifboolexpr{#2==0}{%b=cx
- \EquaBaseL[#1]{#4}{}{}{#3}
+ \EquaBaseL[#1]{#4}{0}{0}{#3}
}{%
\xintifboolexpr{#4==0}{%ax+b=0
- \EquaDeuxL[#1]{#2}{#3}{}{0}
+ \EquaDeuxL[#1]{#2}{#3}{0}{0}
}{%ax+b=cx
\xintifboolexpr{#2==#4}{%
\xintifboolexpr{#3==0}{%ax=ax
@@ -132,7 +132,7 @@
}{}
}{}
\end{align*}
- \ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.}{}
+ %\ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.}{}
}{%ax+b=cx+d avec a<c % Autre cas délicat
\begin{align*}%
\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}&=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\\
@@ -157,16 +157,16 @@
}{}
}{}
\end{align*}
- \ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.}{}%
+ %\ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.}{}%
}%
+ \ifboolKV[ClesEquation]{Solution}{\EcrireSolutionEquation{#2}{#3}{#4}{#5}}{}%
}%
}%
}%
\fi
}%\\
- % \\
-\newcommand{\ResolEquationL}[5][]{%
+\newcommand\ResolEquationL[5][]{%
\useKVdefault[ClesEquation]%
\setKV[ClesEquation]{#1}%
\xintifboolexpr{#2==0}{%
@@ -178,23 +178,23 @@
}%
}%
{%0x+b=cx+d$
- \EquaDeuxL[#1]{#4}{#5}{}{#3}%
+ \EquaDeuxL[#1]{#4}{#5}{0}{#3}%
}%
}{%
\xintifboolexpr{#4==0}{%ax+b=0x+d
- \EquaDeuxL[#1]{#2}{#3}{}{#5}%
+ \EquaDeuxL[#1]{#2}{#3}{0}{#5}%
}
{%ax+b=cx+d$
\xintifboolexpr{#3==0}{%
\xintifboolexpr{#5==0}{%ax=cx
- \EquaTroisL[#1]{#2}{0}{#4}{}%
+ \EquaTroisL[#1]{#2}{0}{#4}{0}%
}%
{%ax=cx+d
- \EquaTroisL[#1]{#4}{#5}{#2}{}%
+ \EquaTroisL[#1]{#4}{#5}{#2}{0}%
}%
}%
{\xintifboolexpr{#5==0}{%ax+b=cx
- \EquaTroisL[#1]{#2}{#3}{#4}{}%
+ \EquaTroisL[#1]{#2}{#3}{#4}{0}%
}%
{%ax+b=cx+d -- ici
\xintifboolexpr{#2==#4}{%
@@ -230,8 +230,7 @@
}{}
}{}
\end{align*}
- \ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.%
- }{}
+ %\ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.}{}
}{%ax+b=cx+d avec a<c % Autre cas délicat
\begin{align*}%
\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}&=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}\\
@@ -255,9 +254,9 @@
}{}
}{}
\end{align*}
- \ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.%
- }{}%
+ %\ifboolKV[ClesEquation]{Solution}{L'équation $\xintifboolexpr{#2==1}{}{\num{#2}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#3>0}{+\num{#3}}{-\num{\fpeval{0-#3}}}=\xintifboolexpr{#4==1}{}{\num{#4}}\useKV[ClesEquation]{Lettre}\xintifboolexpr{#5>0}{+\num{#5}}{-\num{\fpeval{0-#5}}}$ a une unique solution : \opdiv*{\Coeffb}{\Coeffa}{solution}{resteequa}\opcmp{resteequa}{0}$\ifboolKV[ClesEquation]{LettreSol}{\useKV[ClesEquation]{Lettre}=}{}\displaystyle\ifopeq\opexport{solution}{\solution}\num{\solution}\else\ifboolKV[ClesEquation]{Entier}{\SSimplifie{\Coeffb}{\Coeffa}}{\frac{\num{\Coeffb}}{\num{\Coeffa}}}\fi$.}{}%
}%
+ \ifboolKV[ClesEquation]{Solution}{\EcrireSolutionEquation{#2}{#3}{#4}{#5}}{}%
}%
}%
}%