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}
\vspace*{-2em}
\ct{\bf\LARGE{Corrig\'e du devoir de math\'ematiques}}
\bgex
$a=\lp\sqrt{12}-\sqrt{3}\rp^2
=\lp\sqrt{12}\rp^2-2\sqrt{12}\sqrt{3}+\lp\sqrt{3}\rp^2
=12-2\sqrt{36}+3=12-12+3=3$
\bigskip
$b=\dfrac{2}{2+\sqrt{3}}
=\dfrac{2(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}
=2(2-\sqrt{3})$
\bigskip
$c=\dfrac{x+\dfrac32}{x+\dfrac12}-1
=\dfrac{\dfrac{2x+3}2}{\dfrac{2x+1}2}-1
=\dfrac{2x+3}2\tm\dfrac2{2x+1}-1
=\dfrac{2x+3}{2x+1}-\dfrac{2x+1}{2x+1}
=\dfrac2{2x+1}$
\bigskip
$d=\dfrac{x(3x)^3}{9x^2}=\dfrac{x\tm 3^3x^3}{3^2x^2}=3x^2$
\enex
\bgex
$\bgar{l}(E_1):\ (2x-3)(-x+2)=0
\iff
\la\bgar{lll} &2x-3=0 \\ \mbox{ou, } &-x+2=0\enar\right.
\iff
\la\bgar{lll} &x=\dfrac{3}{2} \\ \mbox{ou, } &x=2\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_1=\la \dfrac{3}{2}\,;\,2\ra}$
\medskip
$\bgar{ll}(E_2):\ &(x+2)(3x-2)-(x+2)(x+1)=0
\iff
(x+2)(2x-3)=0\\[.7em]
&\iff\la\bgar{lll} &x+2=0 \\ \mbox{ou, } &2x-3=0\enar\right.
\iff
\la\bgar{lll} &x=-2 \\ \mbox{ou, } &x=\dfrac{3}{2}\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_2=\la -2\,;\, \dfrac{3}{2}\ra}$
\medskip
$\bgar{ll}(E_3):\ &(x^2-9)(3x+7)=0
\iff
\la\bgar{lll} &x^2-9=0 \\ \mbox{ou, } &3x+7=0\enar\right.\\
&\iff
\la\bgar{lll} &x^2=9 \\ \mbox{ou, } &x=-\dfrac{7}{3}\enar\right.
\iff
\la\bgar{ll} x=-\sqrt{9}=3\ \mbox{ou, } x=\sqrt{9}=3 \\ \mbox{ou, } x=-\dfrac{7}{3}\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_3=\la -\dfrac{7}{3}\,;\,-3\,;\,3\ra}$
\medskip
$\bgar{ll}(E_4):\ &(2x-3)(x+6)-(x+6)=0
\iff
(x+6)\Big[(2x-3)-1\Big]=0\\
&\iff
(x+6)\lb 2x-4\rb=0
\iff
\la\bgar{lll} &x+6=0 \\ \mbox{ou, } &2x-4=0\enar\right.
\iff
\la\bgar{lll} &x=-6 \\ \mbox{ou, } &x=2\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_4=\la -6\,;\,2\ra}$
\medskip
$\bgar{ll}(E_5):\ &\dfrac{2}{2x+5}-\dfrac{1}{4x-3}=0
\iff
\dfrac{6x-11}{(2x+5)(4x-3)}=0
\iff
\la\bgar{ll} &6x-11=0 \\ \mbox{et,}&(2x+5)(4x-3)\not=0\enar\right.\\
&\iff
\la\bgar{ll} &x=\dfrac{11}{6} \\
\mbox{et,}&x\not=-\dfrac{5}{2}\ \mbox{et, } x\not=\dfrac{3}{4}\enar\right.
\hspace*{8.8cm}\ul{\mathcal{S}_5=\la \dfrac{11}{6} \ra}\enar$
\medskip
$\bgar{l}(E_6):\ (2x+3)^2=49
\iff
\la\bgar{lll} &2x+3=-7 \\ \mbox{ou, } &2x+3=7\enar\right.
\iff
\la\bgar{lll} &x=-5 \\ \mbox{ou, } &x=2\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_6=\la -5\,;\,2 \ra}$
\medskip
$\bgar{ll}(E_7): &\dfrac{x}{2x+1}=1
\iff
\dfrac{x}{2x+1}-1=0
\iff
\dfrac{x}{2x+1}-\dfrac{2x+1}{2x+1}=0\\
&\iff
\dfrac{-x-1}{2x+1}=0
\iff\la\bgar{ll}
&-x-1=0 \vspd\\
\mbox{et,}\ &2x+1\not=0
\enar\right.
\iff
\la\bgar{ll}
&x=-1 \vspd\\
\mbox{et,}\ &x\not=-\frac{1}{2}
\enar\right.
\enar$,
\hfill
\ul{$\mathcal{S}_7=\la -1\ra$}.
\medskip
$\bgar{ll}(E_8):
&\dfrac{6x(x+1)}{x-4}=30+\dfrac{120}{x-4}
\iff\dfrac{6x(x+1)}{x-4}-\dfrac{30(x-4)}{x-4}-\dfrac{120}{x-4}=0\\
&\iff\dfrac{6x^2+6x-30x+120-120}{x-4}=0
\iff\dfrac{6x^2-24x}{x-4}=0\enar$
On peut, et doit, factoriser le num\'erateur:
$(E_8) \iff \dfrac{6x(x-4)}{x-4}=0$
C'est une \'equation quotient, et donc,
$\la\bgar{ll}6x(x-4)=0 \\\text{et, }x-4\not=0\enar\right.
\iff \la\bgar{ll}6x=0 \text{ ou, } x-4=0\\\text{et,}x-4\not=0\enar\right.$
\hfill
$\ul{\mathcal{S}_8=\la0\ra}$.
\enex
\label{LastPage}
\end{document}
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