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\medskip
%$\bgar{ll}(E_0): &\dfrac{6-x}{2x+6}=\dfrac1{x+1}
%\iff \dfrac{6-x}{2x+6}-\dfrac1{x+1}=0\\[1em]
%&\iff \dfrac{(6-x)(x+1)-(2x+6)}{(2x+6)(x+1)}=0
%\iff \dfrac{-x^2+3x}{(2x+6)(x+1)}=0\\
%&\iff\la\bgar{ll}-x^2+3x=x(-x+3)=0\\\text{et }(2x+6)(x+1)\not=0\enar\right.
%\iff\la\bgar{ll}x=0 \text{ ou } x=3\\\text{et }x\not=-3 \text{ et } x\not=-1\enar\right.
%\enar$
%\hfill$\mathcal{S}=\la0;3\ra$
$\bgar{ll}(E_1):\ &(x^2-11)(3x+7)=0
\iff
\la\bgar{lll} &x^2-11=0 \\ \mbox{ou, } &3x+7=0\enar\right.\\
&\iff
\la\bgar{lll} &x^2=11 \\ \mbox{ou, } &x=-\dfrac{7}{3}\enar\right.
\iff
\la\bgar{ll} x=-\sqrt{11}\ \mbox{ou, } x=\sqrt{11} \\ \mbox{ou, } x=-\dfrac{7}{3}\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_1=\la -\dfrac{7}{3}\,;\,-\sqrt{11}\,;\,\sqrt{11}\ra}$
\medskip
$\bgar{l}(E_2):\ (2x+3)^2=49
\iff
\la\bgar{lll} &2x+3=-7 \\ \mbox{ou, } &2x+3=7\enar\right.
\iff
\la\bgar{lll} &x=-5 \\ \mbox{ou, } &x=2\enar\right.
\enar$
\hfill
$\ul{\mathcal{S}_2=\la -5\,;\,2 \ra}$
\medskip
$\bgar{ll}(E_3):
&3x(2x+1)=2x
\iff3x(2x+1)-2x=0
\iff x\Bigl(3(2x+1)-2\Bigr)=0\\
&\iff x\lp 6x+1\rp=0
\iff\la\bgar{ll}x=0\\x=-\dfrac16\enar\right.
\enar$
\hfill $\mathcal{S}_3=\la0;-\dfrac16\ra$
\medskip
$\bgar{ll}(E_4):\ &\dfrac{2}{2x+5}-\dfrac{1}{4x-3}=0
\iff
\dfrac{6x-11}{(2x+5)(4x-3)}=0
\iff
\la\bgar{ll} &6x-11=0 \\ \mbox{et,}&(2x+5)(4x-3)\not=0\enar\right.\\
&\iff
\la\bgar{ll} &x=\dfrac{11}{6} \\
\mbox{et,}&x\not=-\dfrac{5}{2}\ \mbox{et, } x\not=\dfrac{3}{4}\enar\right.
\hspace*{8.8cm}\ul{\mathcal{S}_4=\la \dfrac{11}{6} \ra}\enar$
\medskip
$\bgar{ll}(E_5): &\dfrac{3x}{2x+1}=2x
\iff
\dfrac{3x}{2x+1}-2x=0
\iff
x\lp\dfrac{3}{2x+1}-2\rp=0
\\
&\iff x\dfrac{-2x+2}{2x+1}=0
\iff\la\bgar{ll}
&x(-4x+1)=0 \vspd\\
\mbox{et,}\ &2x+1\not=0
\enar\right.
\iff
\la\bgar{ll}
x=0 \text{ ou } x=\dfrac14 \vspd\\
\mbox{et,}\ x\not=-\frac{1}{2}
\enar\right.
\enar$,
\hfill
\ul{$\mathcal{S}_5=\la 0 ; \dfrac14\ra$}.
\medskip
$\bgar{ll}(E_6):
&\lp\dfrac2{2x+1}\rp^2=9
\iff\la\bgar{ll}\dfrac2{2x+1}=\sqrt9=3\\[1em]\dfrac2{2x+1}=-\sqrt9=-3\enar\right.
\iff\la\bgar{ll}\dfrac2{2x+1}-3=0\\[1em]\dfrac2{2x+1}+3=0\enar\right.\\
&\iff\la\bgar{ll}\dfrac{-6x-1}{2x+1}=0\\[1em]\dfrac{6x+5}{2x+1}=0\enar\right.
\iff\la\bgar{ll}&-6x-1=0\\\text{ou}&6x+5=0\\\text{et}&2x+1\not=0\enar\right.
\iff\la\bgar{ll}&x=-\dfrac16\\\text{ou}&x=\dfrac56\\\text{et}&x\not=-\dfrac12\enar\right.
\enar$
\hfill $\mathcal{S}_6=\la-\dfrac16;\dfrac56\ra$
\enex
\clearpage
\bgex
\bgen[a)]
\item De mani\`ere g\'en\'erale: $\V{AB}(x_B-x_A;y_B-y_A)$,
et donc
$\V{AB}(8;-10)$ et $\V{AD}(20;1)$
\item On a alors,
$AB=\sqrt{8^2+(-10)^2}=\sqrt{164}=2\sqrt{41}$
et $AD=\sqrt{20^2+1^2}=\sqrt{401}$
\item Soit $F(x;y)$, alors
d'une part $\V{AF}(x+1;y-2)$ \\
et d'autre part, d'après la question précédente, $\V{AB}+\V{AD}$ a pour coordonnées $(28;-9)$. \\
Ainsi,
\[\bgar{ll}\V{AF}=\V{AB}+\V{AD}
&\iff\la\bgar{ll}x+1=28\\y-2=-9\enar\right.\\[1em]
&\iff\la\bgar{ll}x=27\\y=-7\enar\right.
\enar\]
On a donc trouvé $F(27;-7)$.
\item Le centre $I$ est le milieu de $[AB]$, et a pour
coordonn\'ees:
$\dsp I\lp\frac{-1+7}{2};\frac{2+(-8)}{2}\rp$,
soit, \ul{$I(3;-3)$}.
On calcule: $\V{AB}(8;-10)$, et donc,
$AB=\sqrt{8^2+(-10)^2}=\sqrt{164}=\sqrt{4\tm41}=2\sqrt{41}$.
On en d\'eduit que le rayon du cercle est $\dsp R=\frac{AB}{2}=\sqrt{41}$.
\item On calcule la longueur $IE$:
on a $\V{IE}(4;5)$, et donc
$IE=\sqrt{4^2+5^2}=\sqrt{41}=R$.
$E$ est donc bien sur le cercle de diam\`etre $[AB]$.
\item Soit $G(x;y)$ alors
d'une part $\V{AG}(x+1;y-2)$ et d'autre part $\V{GB}(7-x;-8-y)$.\\
On a alors
\[\bgar{ll}\V{AG}=\V{GB}
&\iff\la\bgar{ll}x+1=7-x\\y-2=-8-y\enar\right.\\
&\iff\la\bgar{ll}2x=6\\2y=-6\enar\right.\\
&\iff\la\bgar{ll}x=3\\y=-3\enar\right.
\enar\]
\textsl{Remarque: ce point $G$ est le milieu du segment $[AB]$ (faire un dessin avec les vecteurs $\V{AG}$ et $\V{GB}$) !}
\enen
\enex
\label{LastPage}
\end{document}
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