Source Latex
de la correction du devoir
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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
$a=\dfrac{\dfrac{3}{7}-\dfrac{2}{5}}{\dfrac{3}{5}}
=\dfrac{\dfrac{1}{7\tm5}}{\dfrac{3}{5}}
=\dfrac{1}{7\tm5}\tm\dfrac{5}{3}
=\dfrac{1}{21}$
\bigskip
$b=2-5\lp\dfrac{7}{5}-1\rp^2
=2-5\lp\dfrac{2}{5}\rp^2=2-5\tm\dfrac{2^2}{5^2}
=2-\dfrac{4}{5}=\dfrac{6}{5}$
\enex
\bgex
$A(x)=\dfrac{2}{3x+4}+\dfrac{5}{6x+7}
=\dfrac{2(6x+7)+5(3x+4))}{(3x+4)(6x+7)}
=\dfrac{27x+34}{(3x+4)(6x+7)}$
\bigskip
$B(x)=\dfrac{3x+2}{2x-3}-1=\dfrac{3x+2-(2x-3)}{2x-3}
=\dfrac{x+5}{2x-3}$
\bigskip
$C(x)=\dfrac{1-x^2}{1-2x}-\dfrac{1+2x}{4}
=\dfrac{4\lp1-x^2\rp-(1+2x)(1-2x)}{4(1-2x)}
=\dfrac{4-4x^2-\lp1-4x^2\rp}{4(1-2x)}
=\dfrac{3}{4(1-2x)}$
\enex
\bgex
$A=-6x^2+8x$
\bigskip
$B=x^2+4x+3$
\bigskip
$C=4x^2-4x+1$
\bigskip
$D=2x\lp x^2+2x\dfrac12+\lp\dfrac12\rp^2\rp
=2x\lp x^2+x+\dfrac14\rp
=2x^3+2x^2+\dfrac12x$
\enex
\label{LastPage}
\end{document}
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