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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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\vspace*{-2em}
\ct{\bf\LARGE{Corrig\'e du devoir de math\'ematiques}}
\bgex
\bgen[a)]
\item $\V{AD}=\V{AB}+\V{AC}$
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%
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\rput(2.2,1.8){$\V{AC}$}
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\rput(6,3.5){$\V{AC}$}
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\item $\V{CE}=\V{AC}-\V{AB}=\V{AC}+\V{BA}$
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\item $\V{BF}=\dfrac12\V{AB}-\dfrac32\V{CA}=\dfrac12\V{AB}+\dfrac32\V{AC}$
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\enen
\enex
\clearpage
\bgex
$\bgar{ll}(E_1):\ &(2x+1)(x+3)=(x+3)
\iff(2x+1)(x+3)-(x+3)=0\\
&\iff(x+3)\Bigl((2x+1)-1\Bigr)=0
\iff(2x+1)\Bigl(2x\Bigr)=0\\
&\iff\la\bgar{lll} &2x+1=0 \\ \mbox{ou, } &2x=0\enar\right.
\iff
\la\bgar{lll} &x=-\dfrac12 \\ \mbox{ou, } &x=0\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_1=\la-\dfrac12\,;\,0\ra}$
\medskip
$\bgar{ll}(E_2):\ &\sqrt{2x+1}=6\geqslant0
\iff\la\bgar{lll} &2x+1=6^2=36 \\ \mbox{et, } &2x+6\geqslant0\enar\right.
\iff
\la\bgar{lll} &x=\dfrac{35}2 \\ \mbox{et, } &2\tm\dfrac{35}2+1=36\geqslant0\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_2=\la \dfrac{35}{2}\ra}$
\medskip
$\bgar{ll}(E_3):\ &(2x+1)^2=-9<0\enar$ et donc cette équation n'a aucune solution
\hfill
$\ul{\mathcal{S}_3=\emptyset}$
\medskip
$\bgar{ll}(E_4):\ &\dfrac{x}{2x+1}=\dfrac{2x}{4x-1}
\iff
\dfrac{-3x}{(2x+1)(4x-1)}=0
\iff
\la\bgar{ll} &-3x=0 \\ \mbox{et,}&(2x+1)(4x-1)\not=0\enar\right.\\
&\iff
\la\bgar{ll} &x=0 \\
\mbox{et,}&(2x+1)(4x-1)=-1\not=0\enar\right.\enar$
\hfill$\ul{\mathcal{S}_4=\la0\ra}$
\medskip
$\bgar{ll}(E_5):\ (3x-1)^2=4\geqslant0
\iff
\la\bgar{lll} &3x-1=\sqrt4=2 \\ \mbox{ou, } &3x-1=-\sqrt4=-2\enar\right.
\iff
\la\bgar{lll} &x=1 \\ \mbox{ou, } &x=-\dfrac13\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_5=\la 1\,;\,-\dfrac13\ra}$
\medskip
$\bgar{ll}(E_6):\ &(2x-6)\sqrt{x-5}=0
\iff
\la\bgar{lll} &2x-6=0 \\ \mbox{ou, } &\sqrt{x-5}=0\enar\right.
\iff
\la\bgar{lll} &x=-3 \\ \mbox{ou, } &x-5=0 \text{ et } x-5\geqslant0\enar\right.
\\
&
\iff
\la\bgar{lll} &x=-3 \\ \mbox{ou, } &x=5 \text{ et } x-5\geqslant0\enar\right.
\enar$
Or, pour $x=3$, on a $x-5=-2<0$ et $x=3$ n'est donc pas solution.
\hfill
$\ul{\mathcal{S}_6=\la 5 \ra}$
\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
$\bgar{ll}(E_7):
&\dfrac{6x(x+1)}{x-4}=30+\dfrac{120}{x-4}
\iff\dfrac{6x(x+1)}{x-4}-\dfrac{30(x-4)}{x-4}-\dfrac{120}{x-4}=0\\
&\iff\dfrac{6x^2+6x-30x+120-120}{x-4}=0
\iff\dfrac{6x^2-24x}{x-4}=0\enar$
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
$\ul{\mathcal{S}_8=\la0\ra}$.
$(E_8)$:
$\bgar[t]{ll}&x^2(3x+7)=9(3x+7)
\iff x^2(3x+7)-9(3x+7)=0
\iff \lp x^2-9\rp(3x+7)=0\\
&\iff\la\bgar{lll} &x^2-9=0 \\ \mbox{ou, } &3x+7=0\enar\right.
\iff\la\bgar{lll} &x^2=9 \\ \mbox{ou, } &x=-\dfrac{7}{3}\enar\right.
\iff\la\bgar{ll} x=-\sqrt{9}=3\ \mbox{ou, } x=\sqrt{9}=3 \\ \mbox{ou, } x=-\dfrac{7}{3}\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_8=\la -\dfrac{7}{3}\,;\,-3\,;\,3\ra}$
\enex
\label{LastPage}
\end{document}
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