Source Latex
de la correction du devoir
\documentclass[12pt]{article}
\usepackage{amsfonts}\usepackage{amssymb}
\usepackage[french]{babel}
\usepackage{amsmath}
\usepackage[utf8]{inputenc}
\usepackage{calc}
\usepackage{enumerate}
\usepackage{pst-all}
\usepackage{hyperref}
\hypersetup{
pdfauthor={Yoann Morel},
pdfsubject={Corrigé du devoir de mathématiques: Calcul algébrique, fractions, développement et factorisation},
pdftitle={Corrigé du devoir de mathématiques},
pdfkeywords={calcul algébrique, fraction, développement, factorisation, identités remarquables, mathématiques, seconde, 2nde}
}
\hypersetup{
colorlinks = true,
linkcolor = red,
anchorcolor = red,
citecolor = blue,
filecolor = red,
urlcolor = red
}
\voffset=-1.5cm
% Raccourcis diverses:
\newcommand{\nwc}{\newcommand}
\nwc{\dsp}{\displaystyle}
\nwc{\bge}{\begin{equation}}\nwc{\ene}{\end{equation}}
\nwc{\bgar}{\begin{array}}\nwc{\enar}{\end{array}}
\nwc{\bgit}{\begin{itemize}}\nwc{\enit}{\end{itemize}}
\nwc{\bgen}{\begin{enumerate}}\nwc{\enen}{\end{enumerate}}
\nwc{\la}{\left\{}\nwc{\ra}{\right\}}
\nwc{\lp}{\left(}\nwc{\rp}{\right)}
\nwc{\lb}{\left[}\nwc{\rb}{\right]}
\nwc{\ul}{\underline}
\def\N{{\rm I\kern-.1567em N}}
\def\R{{\rm I\kern-.1567em R}}
%\def\Q{\mathbb{Q}}
\def\Q{{\rm \psline[linewidth=.06em](.1,0.01)(.1,.28) Q}}
\def\Z{{\sf Z\kern-4.5pt Z}}
\nwc{\tm}{\times}
\newcounter{nex}\setcounter{nex}{0}
\newenvironment{EX}{%
\stepcounter{nex}
\bigskip\noindent{\large {\bf Exercice }\arabic{nex}}\hspace{0.5cm}
}{}
\nwc{\bgex}{\begin{EX}}\nwc{\enex}{\end{EX}}
\newcommand{\ct}{\centerline}
\nwc{\bgmp}{\begin{minipage}}\nwc{\enmp}{\end{minipage}}
% Dimensions des pages
\textwidth=18.5cm
\textheight=26.cm
\topmargin=-1.2cm
\oddsidemargin=-1.4cm
\footskip=.8cm
% Bandeau en bas de page
\newcommand{\TITLE}{Correction du devoir de mathématiques}
\author{Y. Morel}
\date{}
\usepackage{fancyhdr}
\pagestyle{fancyplain}
\setlength{\headheight}{0cm}
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\footrulewidth}{0.1pt}
\lhead{}\chead{}\rhead{}
\lfoot{Y. Morel - \url{http://xymaths.free.fr/Lycee/2nde/}}
\rfoot{\TITLE\ - \thepage/\pageref{LastPage}}
\cfoot{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\textbf{\fbox{\Large A}}\hfill{\Large\textbf\TITLE}\hfill \,
\bigskip
\bgex
\[A=\dfrac{2x}{x+3}-2
=\dfrac{2x}{x+3}-\dfrac{2(x+3)}{x+3}
=\dfrac{2x-2(x+3)}{x+3}=\dfrac{-6}{x+3}\]
\[\bgar{ll}
B&=\dfrac{1}{2x+1}-\dfrac{1}{2x-1}
=\dfrac{1(2x-1)}{(2x+1)(2x-1)}-\dfrac{1(2x+1)}{(2x+1)(2x-1)}\\[1.2em]
&=\dfrac{(2x-1)-(2x+1)}{(2x+1)(2x-1)}
=\dfrac{-2}{(2x+1)(2x-1)}\enar\]
\[\bgar{ll}
C&=2+\dfrac{\dfrac12\sqrt2}{1+\dfrac{\sqrt2}{2}}
=2+\dfrac{\dfrac{\sqrt2}{2}}{\dfrac{2+\sqrt2}{2}}
=2+\dfrac{\sqrt2}{2}\tm\dfrac{2}{2+\sqrt2}
=2+\dfrac{\sqrt2}{2+\sqrt2}\\[2.4em]
&=2+\dfrac{\sqrt2(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)}
=2+\dfrac{2\sqrt2-2}{2^2-\sqrt2^2}
=2+\dfrac{2(\sqrt2-1)}{2}\\[1.5em]
&=2+\sqrt2-1=1+\sqrt2
\enar\]
\[\bgar{ll}
D&=\lp3-\dfrac{5}{\sqrt2+2}\rp^2
=\lp 3-\dfrac{5(\sqrt2-2)}{(\sqrt2+2)(\sqrt2-2)}\rp^2
=\lp 3-\dfrac{5\sqrt2-10}{-2}\rp^2\\[1.6em]
&=\lp\dfrac{3\tm(-2)-\lp5\sqrt2-10\rp}{-2}\rp^2
=\lp\dfrac{4-5\sqrt2}{-2}\rp^2
=\dfrac{\lp4-5\sqrt2\rp^2}{(-2)^2}\\[1.6em]
&=\dfrac{4^2-2\tm4\tm5\sqrt2+\lp5\sqrt2\rp^2}{4}
=\dfrac{16-40\sqrt2+50}{4}
=\dfrac{66-40\sqrt2}{4}\\[1.6em]
&=\dfrac{2\lp33-20\sqrt2\rp}{2\tm2}
=\dfrac{33-20\sqrt2}{2}
\enar\]
\enex
\bgex
\[A=(2x+1)-(2-x)(2x+1)
=(2x+1)\Bigl( 1-(2-x)\Bigr)
=(2x+1)(-1+x)\]
\[B=(x+3)(-x+1)-(x+3)^2
=(x+3)\Bigl((-x+1)-(x+3)\Bigr)
=(x+3)(-2x-2)\]
\[C=(2x+1)^2-4=(2x+1)^2-2^2
=\Bigl((2x+1)-2\Bigr)\Bigl((2x+1)+2\Bigr)
=(2x-1)(2x+3)\]
\enex
\label{LastPage}
\end{document}
Télécharger le fichier source