Source Latex
de la correction du devoir
\documentclass[11pt,onecolumn,a4paper]{article}
\usepackage[french]{babel}
%%\selectlanguage{francais}
\usepackage[utf8]{inputenc}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{epsf}
\usepackage{enumerate}
\usepackage{array}
\usepackage{pst-all}
\usepackage{hyperref}
\hypersetup{
pdfauthor={Yoann Morel},
pdfsubject={Devoir de mathématiques de seconde},
pdftitle={Devoir de mathématiques de 2nde},
pdfkeywords={puissances, calcul algébrique, fractions, développer, factoriser, identités remarquables}
}
\hypersetup{
colorlinks = true,
linkcolor = blue,
anchorcolor = red,
citecolor = blue,
filecolor = red,
urlcolor = red
}
\voffset=-1cm
% 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}
\nwc{\tm}{\times}
\nwc{\V}{\overrightarrow}
\newcommand{\zb}{\mbox{$0\hspace{-0.67em}\mid$}}
\newcommand{\db}{\mbox{$\hspace{0.1em}|\hspace{-0.67em}\mid$}}
\newcommand{\ct}{\centerline}
\nwc{\bgsk}{\bigskip}
\nwc{\vsp}{\vspace{0.1cm}}
\nwc{\vspd}{\vspace{0.2cm}}
\nwc{\vspt}{\vspace{0.3cm}}
\nwc{\vspq}{\vspace{0.4cm}}
\def\N{{\rm I\kern-.1567em N}}
\def\D{{\rm I\kern-.1567em D}}
\def\R{{\rm I\kern-.1567em R}}
\def\C{{\rm C\kern-4.7pt
\vrule height 7.7pt width 0.4pt depth -0.5pt \phantom {.}}}
\def\Q{\mathbb{Q}}
\def\Z{{\sf Z\kern-4.5pt Z}}
\def\euro{\mbox{\raisebox{.25ex}{{\it =}}\hspace{-.5em}{\sf C}}}
\newcounter{nex}[section]\setcounter{nex}{0}
\newenvironment{EX}{%
\stepcounter{nex}
\bgsk{\noindent{{\bf Exercice }}\arabic{nex}}\hspace{0.5cm}
}{}
\nwc{\bgex}{\begin{EX}}\nwc{\enex}{\end{EX}}
\nwc{\bgfg}{\begin{figure}}\nwc{\enfg}{\end{figure}}
\nwc{\epsx}{\epsfxsize}\nwc{\epsy}{\epsfysize}
\nwc{\bgmp}{\begin{minipage}}\nwc{\enmp}{\end{minipage}}
\newenvironment{centerpage}{\vspace*{\fill}}{
\protect\vspace*{\fill}}
\setlength{\columnsep}{30pt} % default=10pt
\setlength{\columnseprule}{1pt} % default=0pt (no line)
\setlength{\headsep}{0in} % default=0.35in
\setlength{\parskip}{0ex}
\setlength{\parindent}{0mm}
\voffset=-1cm
\textheight=26.8cm
\textwidth=18.5cm
\topmargin=0cm
\headheight=-0.cm
\footskip=1.cm
\oddsidemargin=-1.cm
\usepackage{fancyhdr}
\pagestyle{fancyplain}
\setlength{\headheight}{0cm}
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\footrulewidth}{.1pt}
\lfoot{Y. Morel - \href{https://xymaths.fr/Lycee/2nde/Mathematiques-2nde.php}{xymaths - 2nde}}
\cfoot{}
\rfoot{Corrigé du devoir de mathématiques - \thepage/\pageref{LastPage}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\ct{\bf\LARGE{Corrig\'e du devoir de math\'ematiques}}
\bgex
$A(x)=\lp1+3x\rp^2-\lp1-3x\rp^2=1+6x+9x^2-\lp1-6x+9x^2\rp=12x$
$B(x)=\lp 2x+3\rp\lp x+1\rp+\lp2x-1\rp^2=2x^2+2x+3x+3+4x^2-4x+1=6x^2+x+4$
\enex
\bgex
$C(x)=(2x-2)(2x-6)-(x-3)(2x-6)=(2x-6)\Bigl[ (2x-2)-(x-3)\Bigr]=(2x-6)(x+1)$
$D(x)=(-x+3)(2x-1)+(-x+3)=(-x+3)\Bigl[ (2x-1)+1 \Bigr]=(-x+3)(2x)$
\enex
\bgex
$A(x)=\dfrac{2}{3x+4}-\dfrac{5}{6x+7}
=\dfrac{2(6x+7)-5(3x+4))}{(3x+4)(6x+7)}
=\dfrac{-3x-6}{(3x+4)(6x+7)}$
\medskip
$B(x)=\dfrac{3x+2}{2x-3}-1=\dfrac{3x+2-(2x-3)}{2x-3}
=\dfrac{x+5}{2x-3}$
\enex
\bgex
\[
\bullet\ \ a=\lp\sqrt{12}-\sqrt{3}\rp^2
=\lp\sqrt{12}\rp^2-2\sqrt{12}\sqrt{3}+\lp\sqrt{3}\rp^2
=12-2\sqrt{36}+3=12-12+3=3
\]
\[
\bullet\ \ b=(3\sqrt{2})^2-(\sqrt{2}-1)^2
=3^2\sqrt{2}^2-\lp \sqrt{2}^2-2\sqrt{2}+1^2\rp
=18-\lp 2-2\sqrt{2}+1\rp
=15+2\sqrt{2}
\]
\[
\bullet\ \ c=\frac{15}{\sqrt{5}}=\frac{15\tm\sqrt{5}}{\sqrt{5}\tm\sqrt{5}}
=\frac{15\sqrt{5}}{5}=3\sqrt{5}
\hspace{1cm}
\bullet\ \ d=\frac{2}{2+\sqrt{3}}
=\frac{2(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}
=2(2-\sqrt{3})
\]
\[
\bullet\ \ e=\frac{\sqrt{3}+\sqrt{12}}{\sqrt{3}-\sqrt{12}}
=\frac{(\sqrt{3}+\sqrt{12})(\sqrt{3}+\sqrt{12})}
{(\sqrt{3}-\sqrt{12})(\sqrt{3}+\sqrt{12})}
=\frac{(\sqrt{3})^2+2\sqrt{3}\sqrt{12}+(\sqrt{12})^2}{3-12}
=\frac{27}{-9}=-3
\]
\enex
\bgex
$A=(a^{-3}b^{4})^3$ et $B=a^3b^{-2}$, alors
\vspt
$\bullet\ \ A=a^{-3\tm3}b^{4\tm3}=a^{-9}b^{12}$
\hspace{1cm}
$\bullet\ \ A\tm B=a^{-9}b^{12}a^3b^{-2}=a^{-6}b^{10}$
\hspace{1cm}
$\dsp\bullet\ \ \frac{A}{B}=\frac{a^{-9}b^{12}}{a^3b^{-2}}
=a^{-12}b^{14}$
\enex
\label{LastPage}
\end{document}
Télécharger le fichier source 
