Source Latex
de la correction du devoir
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% %
% Generateur automatique de devoir, %
% par Y. Morel %
% http://xymaths.free.fr %
% %
% Genere le: %
% Thursday 30 November 2017 %
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\begin{document}
\ifthenelse{\pageref{LastPage}=1}
{\pagestyle{empty}}%
{%
\lfoot{}\cfoot{}\rfoot{\thepage/\pageref{LastPage}}}
\ct{\bf\LARGE{Corrig\'e du devoir de math\'ematiques}}
\bgen[$(E_1)$]
\item
$5x-\bigl[ 4 - (3x-2)\bigr]=x+8
\iff 5x-\bigl[6-3x\bigr]=x+8
\iff 8x-6=x+8
\iff 7x=14
\iff x=2$
Donc, $\mathcal{S}=\la 2\ra$.
\item $(2x-3)(-x+2)=0
\iff \la\bgar{lll} &2x-3=0 \\ \mbox{ou, } &-x+2=0\enar\right.
\iff \la\bgar{lll} &x=\dfrac{3}{2} \\ \mbox{ou, } &x=2\enar\right.$
donc, $\mathcal{S}=\la \dfrac{3}{2}\,;\,2\ra$
\item $x(x+3)=2(x+3)
\iff x(x+3)-2(x+3)=0
\iff (x+3)\lb x-2\rb=0$. \\
C'est une \'equation produit et donc
$\la\bgar{ll}
&x+3=0\\
\mbox{ou, }\ &x-2=0
\enar\right.
\iff
\la\bgar{ll}
&x=-3\\
\mbox{ou, }\ &x=2
\enar\right.$.
Donc, $\mathcal{S}=\la -3;2\ra$.
\item $(x^2-5)(3x+7)=0
\iff
\la\bgar{lll} &x^2-5=0 \\ \mbox{ou, } &3x+7=0\enar\right.
\iff
\la\bgar{lll} &x^2=5 \\ \mbox{ou, } &x=-\dfrac{7}{3}\enar\right.
\iff
\la\bgar{ll} x=-\sqrt{5}\ \mbox{ou, } x=\sqrt{5} \\ \mbox{ou, } x=-\dfrac{7}{3}\enar\right.
$
donc,
$\mathcal{S}=\la -\dfrac{7}{3}\,;\,-\sqrt{5}\,;\,\sqrt{5}\ra$
\item $(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.
$
donc,
$\mathcal{S}=\la -6\,;\,2\ra$
\item $\dfrac{x^2-25}{2x-10}=0
\iff
\la\bgar{ll} &x^2-25=0 \\ \mbox{et, }&2x-10\not=0\enar\right.
\iff
\la\bgar{ll} &x^2=25 \\ \mbox{et, }&x\not=5\enar\right.
\iff
\la\bgar{ll} x=-5\ \mbox{ou,}\ x=5 \\ \mbox{et, } x\not=5\enar\right.
$
donc,
$\mathcal{S}=\la -5\ra$
\item $\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.
$
donc,
$\mathcal{S}=\la \dfrac{11}{6} \ra$
\item $(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.
$
donc,
$\mathcal{S}=\la -5\,;\,2 \ra$
\enen
\label{LastPage}
\end{document}
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