Source Latex
de la correction du devoir
\documentclass[12pt]{article}
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\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}
}
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% Bandeau en bas de page
\newcommand{\TITLE}{Corrigé du devoir de mathématiques}
\author{Y. Morel}
\date{}
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\lfoot{Y. Morel - \url{http://xymaths.free.fr/Lycee/2nde/}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
{\bf\fbox{\Large B}}\hfill{\bf\LARGE{\TITLE}}\hfill\,
\bgex
$A=\dfrac{3(x+1)-(x+2)}{(x+2)(x+1)}=\dfrac{2x+1}{(x+2)(x+1)}$
$B=\dfrac{3x-3(x+2)}{x+2}=\dfrac{-6}{x+2}$
$C=\dfrac{(3x-1)-(3x+1)}{(3x+1)(3x-1)}=\dfrac{-2}{9x^2-1}$
$D=3+\dfrac{\dfrac13x}{\dfrac{3+x}{3}}=3+\dfrac13x\tm\dfrac{3}{3+x}
=3+\dfrac{x}{3+x}=\dfrac{3(3+x)+x}{3+x}=\dfrac{4x+9}{x+3}$
$E=\lp\dfrac{2(x+3)-5}{x+3}\rp^2=\lp\dfrac{2x+1}{x+3}\rp^2
=\dfrac{(2x+1)^2}{(x+3)^2}=\dfrac{4x^2+4x+1}{x^2+6x+9}$
\enex
\bgex
$A=x^2+5x+5$
\qquad
$B=x^2+4x+4$
\qquad
$C=4x^2-4x+1$
\\
$D=3x\lp x^2+2x\dfrac13+\lp\dfrac13\rp^2\rp
=3x\lp x^2+\dfrac{2x}{3}+\dfrac19\rp
=3x^3+2x^2+\dfrac13x$
\enex
\bgex
$A=(x+2)\Bigl( (3x+2)-(2x+1)\Bigr)=(x+2)(x+1)$
$B=(3x+1)\Bigl( 1-(4-x)\Bigr)=(3x+1)(-3+x)$
$C=(x+2)\Bigl((-x+1)-(x+2)\Bigr)=(x+2)(-2x-1)$
\enex
\label{LastPage}
\end{document}
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