Source Latex
sujet 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}
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{%
\lfoot{}\cfoot{}\rfoot{\thepage/\pageref{LastPage}}}
\ct{\bf\LARGE{Devoir de math\'ematiques}}
\medskip
R\'esoudre les \'equations:
$(E_1)\ 5x-\bigl[ 4 - (3x-2)\bigr]=x+8$
\qquad $(E_2)\ (2x-3)(-x+2)=0$\\[1.2em]
$(E_3)\ x(x+3)=2(x+3)$
\qquad $(E_4)\ (x^2-5)(3x+7)=0$
\qquad $(E_5)\ (2x-3)(x+6)-(x+6)=0$
$(E_6)\ \dfrac{x^2-25}{2x-10}=0$
\qquad $(E_7)\ \dfrac{2}{2x+5}-\dfrac{1}{4x-3}=0$
\qquad $(E_8)\ (2x+3)^2=49$
\bigskip
\hrulefill
\bigskip
\ct{\bf\LARGE{Devoir de math\'ematiques}}
\medskip
R\'esoudre les \'equations:
$(E_1)\ 5x-\bigl[ 4 - (3x-2)\bigr]=x+8$
\qquad $(E_2)\ (2x-3)(-x+2)=0$\\[1.2em]
$(E_3)\ x(x+3)=2(x+3)$
\qquad $(E_4)\ (x^2-5)(3x+7)=0$
\qquad $(E_5)\ (2x-3)(x+6)-(x+6)=0$
$(E_6)\ \dfrac{x^2-25}{2x-10}=0$
\qquad $(E_7)\ \dfrac{2}{2x+5}-\dfrac{1}{4x-3}=0$
\qquad $(E_8)\ (2x+3)^2=49$
\bigskip
\hrulefill
\bigskip
\ct{\bf\LARGE{Devoir de math\'ematiques}}
\medskip
R\'esoudre les \'equations:
$(E_1)\ 5x-\bigl[ 4 - (3x-2)\bigr]=x+8$
\qquad $(E_2)\ (2x-3)(-x+2)=0$\\[1.2em]
$(E_3)\ x(x+3)=2(x+3)$
\qquad $(E_4)\ (x^2-5)(3x+7)=0$
\qquad $(E_5)\ (2x-3)(x+6)-(x+6)=0$
$(E_6)\ \dfrac{x^2-25}{2x-10}=0$
\qquad $(E_7)\ \dfrac{2}{2x+5}-\dfrac{1}{4x-3}=0$
\qquad $(E_8)\ (2x+3)^2=49$
\bigskip
\hrulefill
\bigskip
\ct{\bf\LARGE{Devoir de math\'ematiques}}
\medskip
R\'esoudre les \'equations:
$(E_1)\ 5x-\bigl[ 4 - (3x-2)\bigr]=x+8$
\qquad $(E_2)\ (2x-3)(-x+2)=0$\\[1.2em]
$(E_3)\ x(x+3)=2(x+3)$
\qquad $(E_4)\ (x^2-5)(3x+7)=0$
\qquad $(E_5)\ (2x-3)(x+6)-(x+6)=0$
$(E_6)\ \dfrac{x^2-25}{2x-10}=0$
\qquad $(E_7)\ \dfrac{2}{2x+5}-\dfrac{1}{4x-3}=0$
\qquad $(E_8)\ (2x+3)^2=49$
\bigskip
\hrulefill
\bigskip
\ct{\bf\LARGE{Devoir de math\'ematiques}}
\medskip
R\'esoudre les \'equations:
$(E_1)\ 5x-\bigl[ 4 - (3x-2)\bigr]=x+8$
\qquad $(E_2)\ (2x-3)(-x+2)=0$\\[1.2em]
$(E_3)\ x(x+3)=2(x+3)$
\qquad $(E_4)\ (x^2-5)(3x+7)=0$
\qquad $(E_5)\ (2x-3)(x+6)-(x+6)=0$
$(E_6)\ \dfrac{x^2-25}{2x-10}=0$
\qquad $(E_7)\ \dfrac{2}{2x+5}-\dfrac{1}{4x-3}=0$
\qquad $(E_8)\ (2x+3)^2=49$
\bigskip
\hrulefill
\bigskip
\ct{\bf\LARGE{Devoir de math\'ematiques}}
\medskip
R\'esoudre les \'equations:
$(E_1)\ 5x-\bigl[ 4 - (3x-2)\bigr]=x+8$
\qquad $(E_2)\ (2x-3)(-x+2)=0$\\[1.2em]
$(E_3)\ x(x+3)=2(x+3)$
\qquad $(E_4)\ (x^2-5)(3x+7)=0$
\qquad $(E_5)\ (2x-3)(x+6)-(x+6)=0$
$(E_6)\ \dfrac{x^2-25}{2x-10}=0$
\qquad $(E_7)\ \dfrac{2}{2x+5}-\dfrac{1}{4x-3}=0$
\qquad $(E_8)\ (2x+3)^2=49$
\label{LastPage}
\end{document}
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