From 0ef3dfaccca426fb0f3b06fc106ef9b5b0423dad Mon Sep 17 00:00:00 2001 From: Norbert Preining Date: Mon, 23 Aug 2021 03:00:54 +0000 Subject: CTAN sync 202108230300 --- .../contrib/profcollege/latex/PfCEquationPose1.tex | 49 +++++++++++----------- 1 file changed, 24 insertions(+), 25 deletions(-) (limited to 'macros/latex/contrib/profcollege/latex/PfCEquationPose1.tex') 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 a0}{+\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 a0}{+\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}}{}% }% }% }% -- cgit v1.2.3