Source Latex: Corrigé du devoir de mathématiques en seconde


Fichier
Type: Corrigé de devoir
File type: Latex, tex (source)
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Description
Devoir corrigé de mathématiques, 2nde: Calcul algébrique, fractions, développement, identités remarquables, factorisation, racines carrées
Niveau
seconde
Mots clé
devoir corrigé de mathématiques, calcul algébrique, fraction, développement, expression algébrique développée et factorisée, identitées remarquables, racines carrées, maths
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Source Latex de la correction du devoir

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\usepackage{hyperref}
\hypersetup{
    pdfauthor={Yoann Morel},
    pdfsubject={Correction du devoir de mathématiques de seconde},
    pdftitle={Corrigé du devoir de mathématiques de 2nde},
    pdfkeywords={calcul algébrique, fractions, développer, factoriser, identités remarquables}
}
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\lfoot{Y. Morel - \href{https://xymaths.fr/Lycee/2nde/Mathematiques-2nde.php}{xymaths - 2nde}}
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\rfoot{Correction du devoir de mathématiques - \thepage/\pageref{LastPage}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

\ct{\bf\LARGE{Corrig\'e du devoir de math\'ematiques}}

\bgex 
\bigskip

$A(x)=(x+3)(2x-1)-(x+3)(x+2)
=(x+3)\Bigl[(2x-1)-(x+2)\Bigr]
=(x+3)(x-3)$ 

\bigskip
$B(x)=(2x+1)^2-2x(2x+1)
=(2x+1)\Bigl[(2x+1)-2x\Bigr]
=(2x+1)$


\bigskip
$C(x)=(2x+1)+(x+2)(2x+1)
=(2x+1)\Bigl(1+(x+2)\Bigr]
=(2x+1)(x+3)$

\bigskip
$D(x)=(x+3)^2-4
=(x+3)^2-2^2
=(x+3-2)(x+3+2)
=(x+1)(x+5)$
\enex

\bigskip

\bgex
\bigskip

$a=\dfrac{3x+2}{2x-3}-1=\dfrac{3x+2-(2x-3)}{2x-3}
=\dfrac{x+5}{2x-3}$

\bigskip
$b=\dfrac{x+\dfrac32}{x+\dfrac12}-1
=\dfrac{\dfrac{2x+3}2}{\dfrac{2x+1}2}-1
=\dfrac{2x+3}2\tm\dfrac2{2x+1}-1
=\dfrac{2x+3}{2x+1}-\dfrac{2x+1}{2x+1}
=\dfrac2{2x+1}$

\bigskip
$c=\dfrac{15}{\sqrt{5}}=\dfrac{15\tm\sqrt{5}}{\sqrt{5}\tm\sqrt{5}}
=\dfrac{15\sqrt{5}}{5}=3\sqrt{5}$

\bigskip
$d=\lp\sqrt{12}-\sqrt{3}\rp^2
=\lp\sqrt{12}\rp^2-2\sqrt{12}\sqrt{3}+\lp\sqrt{3}\rp^2
=12-2\sqrt{36}+3=12-12+3=3$

\bigskip
$e=(3\sqrt{2})^2-(\sqrt{2}-1)^2
=3^2\sqrt{2}^2-\lp \sqrt{2}^2-2\sqrt{2}+1^2\rp
=18-\lp 2-2\sqrt{2}+1\rp
=15+2\sqrt{2}$


\bigskip
$f=\dfrac{2}{2+\sqrt{3}}
=\dfrac{2(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}
=2(2-\sqrt{3})$

\bigskip
%$g=\dfrac{\sqrt{3}+\sqrt{12}}{\sqrt{3}-\sqrt{12}}
%=\dfrac{(\sqrt{3}+\sqrt{12})(\sqrt{3}+\sqrt{12})}
%{(\sqrt{3}-\sqrt{12})(\sqrt{3}+\sqrt{12})}
%=\dfrac{(\sqrt{3})^2+2\sqrt{3}\sqrt{12}+(\sqrt{12})^2}{3-12}
%=\dfrac{27}{-9}=-3$

\bigskip
%$h=\dfrac{x(3x)^3}{9x^2}=\dfrac{x\tm 3^3x^3}{3^2x^2}=3x^2$
$g=(4x+12)\dfrac{\dfrac{x^2-16}{x+3}}{x-4}
=(4x+12)\dfrac{x^2-16}{x+3}\tm\dfrac1{x-4}
=4(x+3)\dfrac{(x-4)(x+4)}{x+3}\tm\dfrac1{x-4}
=4(x+4)=4x+16$
\enex

\bigskip



\label{LastPage}
\end{document}

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