213 lines
7.1 KiB
TeX
213 lines
7.1 KiB
TeX
|
\documentclass[a4paper,12pt, table]{/media/documents/Cours/Prof/Enseignements/2014-2015/tools/style/classDS}
|
||
|
|
\usepackage{/media/documents/Cours/Prof/Enseignements/2014-2015/2014_2015}
|
||
|
|
%\geometry{left=10mm,right=10mm, top=10mm, bottom=10mm}
|
||
|
|
|
||
|
|
% Title Page
|
||
|
|
\titre{4}
|
||
|
|
% \seconde \premiereS \PSTMG \TSTMG
|
||
|
|
\classe{\seconde}
|
||
|
|
\date{6 mai 2015}
|
||
|
|
%\duree{1 heure}
|
||
|
|
\sujet{6}
|
||
|
|
% DS DSCorr DM DMCorr Corr
|
||
|
|
\typedoc{DM}
|
||
|
|
|
||
|
|
%\printanswers
|
||
|
|
|
||
|
|
\begin{document}
|
||
|
|
|
||
|
|
\maketitle
|
||
|
|
|
||
|
|
Vous devez rendre le sujet avec la copie.
|
||
|
|
|
||
|
|
\begin{questions}
|
||
|
|
|
||
|
|
\question
|
||
|
|
\begin{parts}
|
||
|
|
\part Développer et simplifier les expressions suivantes
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
\begin{subparts}
|
||
|
|
\begin{multicols}{2}
|
||
|
|
\subpart $A = ( -10 x - 7 ) ( -10 - ( -6 x ) )$
|
||
|
|
\begin{solution}
|
||
|
|
|
||
|
|
\begin{eqnarray*}
|
||
|
|
A & = & ( -10 x - 7 ) ( -10 - ( -6 x ) ) \\
|
||
|
|
A & = & ( - 10 x - 7 ) ( -10 - ( - 6 x ) ) \\
|
||
|
|
A & = & ( - 10 x - 7 ) ( - 10 - ( - 6 x ) ) \\
|
||
|
|
A & = & - 10 x - 7 ( - 10 + 6 x ) \\
|
||
|
|
A & = & - 60 x^{ 2 } + 58 x + 70
|
||
|
|
\end{eqnarray*}
|
||
|
|
\end{solution}
|
||
|
|
\subpart $B = ( -1 x - 9 )^{ 2 } + 9$
|
||
|
|
\begin{solution}
|
||
|
|
|
||
|
|
\begin{eqnarray*}
|
||
|
|
A & = & ( -1 x - 9 )^{ 2 } + 9 \\
|
||
|
|
A & = & ( - 1 x - 9 )^{ 2 } + 9 \\
|
||
|
|
A & = & ( - x - 9 )^{ 2 } + 9 \\
|
||
|
|
A & = & ( - x - 9 )^{ 2 } + 9 \\
|
||
|
|
A & = & - x - 9^{ 2 } + 9 \\
|
||
|
|
A & = & - x - 9 ( - x - 9 ) + 9 \\
|
||
|
|
A & = & -1 \times ( -1 ) x^{ 2 } + ( -9 \times ( -1 ) - 1 \times ( -9 ) ) x - 9 \times ( -9 ) + 9 \\
|
||
|
|
A & = & x^{ 2 } + ( 9 + 9 ) x + 81 + 9 \\
|
||
|
|
A & = & x^{ 2 } + 18 x + 90
|
||
|
|
\end{eqnarray*}
|
||
|
|
\end{solution}
|
||
|
|
\subpart $C = 8 x - 7 + 4 ( 4 x - 7 )^{ 2 }$
|
||
|
|
\begin{solution}
|
||
|
|
|
||
|
|
\begin{eqnarray*}
|
||
|
|
A & = & 8 x - 7 + 4 ( 4 x - 7 )^{ 2 } \\
|
||
|
|
A & = & 8 x - 7 + 4 ( 4 x - 7 )^{ 2 } \\
|
||
|
|
A & = & 8 x - 7 + 4 \times 4 x - 7^{ 2 } \\
|
||
|
|
A & = & 8 x - 7 + 4 \times 4 x - 7^{ 2 } \\
|
||
|
|
A & = & 8 x - 7 + 4 \times 4 x - 7 ( 4 x - 7 ) \\
|
||
|
|
A & = & 8 x - 7 + 4 ( 4 \times 4 x^{ 2 } + ( -7 \times 4 + 4 \times ( -7 ) ) x - 7 \times ( -7 ) ) \\
|
||
|
|
A & = & 8 x - 7 + 4 ( 16 x^{ 2 } + ( -28 - 28 ) x + 49 ) \\
|
||
|
|
A & = & 8 x - 7 + 4 ( 16 x^{ 2 } - 56 x + 49 ) \\
|
||
|
|
A & = & 8 x - 7 + 4 ( 16 x^{ 2 } - 56 x + 49 ) \\
|
||
|
|
A & = & 8 x - 7 + 4 \times 16 x^{ 2 } + 4 \times ( -56 ) x + 4 \times 49 \\
|
||
|
|
A & = & 64 x^{ 2 } - 216 x + 189
|
||
|
|
\end{eqnarray*}
|
||
|
|
\end{solution}
|
||
|
|
\end{multicols}
|
||
|
|
\end{subparts}
|
||
|
|
|
||
|
|
\part Factoriser les expressions suivantes
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
\begin{subparts}
|
||
|
|
\begin{multicols}{2}
|
||
|
|
\subpart $A = 5 x^{ 2 } - x$
|
||
|
|
\subpart $B = 81 x^{ 2 } + 100 + 180 x$
|
||
|
|
\subpart $C = 9 x^{ 2 } - 9$
|
||
|
|
\subpart $D = 25 x^{ 2 } - 70 x + 49$
|
||
|
|
|
||
|
|
\end{multicols}
|
||
|
|
\end{subparts}
|
||
|
|
|
||
|
|
\part Résoudre les équations suivantes
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
\begin{subparts}
|
||
|
|
\begin{multicols}{2}
|
||
|
|
\subpart $5 x - 1 = 0$
|
||
|
|
\subpart $- 10 x - 4 = - 4 x + 3$
|
||
|
|
|
||
|
|
\columnbreak
|
||
|
|
|
||
|
|
\subpart $9 x^{ 2 } - 9 x - 1 = 9x^2$
|
||
|
|
\subpart $( -10 x + 8 ) ( -5 x - 5 ) = 0$
|
||
|
|
\end{multicols}
|
||
|
|
\end{subparts}
|
||
|
|
|
||
|
|
\end{parts}
|
||
|
|
|
||
|
|
\question
|
||
|
|
\begin{parts}
|
||
|
|
|
||
|
|
\part Soit $A(-5 ; 3)$, $B(-10 ; -2)$, $C(1 ; 1)$ et $D(-3 ; -3)$. Est-ce que les vecteurs $\vec{AB}$ et $\vec{CD}$ sont colinéaires?
|
||
|
|
|
||
|
|
\part Soit $A(-7 ; 9)$, $B(-4 ; -1)$, $C(7 ; -5)$ et $D(6 ; -2)$. Est-ce que les vecteurs $\vec{AB}$ et $\vec{CD}$ sont colinéaires?
|
||
|
|
|
||
|
|
\part Soit $A(3 ; 10)$, $B(6 ; 1)$, $C(2 ; -6)$ et $D(-1 ; 3)$. Est-ce que les droites $(AC)$ et $(BD)$ sont colinéaires?
|
||
|
|
\end{parts}
|
||
|
|
|
||
|
|
\question
|
||
|
|
\begin{parts}
|
||
|
|
\part Faire les calculs suivants en \textbf{détaillant les étapes} et en simplifiant les fractions.
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
\begin{multicols}{2}
|
||
|
|
\begin{subparts}
|
||
|
|
\subpart $\displaystyle A = \frac{ 10 }{ -4 } \times ( -7 )$
|
||
|
|
\begin{solution}
|
||
|
|
|
||
|
|
\begin{eqnarray*}
|
||
|
|
A & = & \frac{ 10 }{ -4 } \times ( -7 ) \\
|
||
|
|
A & = & \frac{ 10 \times 7 \times ( -1 ) }{ 4 \times ( -1 ) } \\
|
||
|
|
A & = & \frac{ 70 \times ( -1 ) }{ -4 } \\
|
||
|
|
A & = & \frac{ -70 }{ -4 } \\
|
||
|
|
A & = & \frac{ 70 }{ 4 } \\
|
||
|
|
A & = & \frac{ 35 \times 2 }{ 2 \times 2 } \\
|
||
|
|
A & = & \frac{ 35 }{ 2 }
|
||
|
|
\end{eqnarray*}
|
||
|
|
\end{solution}
|
||
|
|
\subpart $\displaystyle B = \frac{ 1 }{ -9 } + \frac{ 5 }{ -45 }$
|
||
|
|
\begin{solution}
|
||
|
|
|
||
|
|
\begin{eqnarray*}
|
||
|
|
A & = & \frac{ 1 }{ -9 } + \frac{ 5 }{ -45 } \\
|
||
|
|
A & = & \frac{ 1 \times 5 }{ -9 \times 5 } + \frac{ 5 \times 1 }{ -45 \times 1 } \\
|
||
|
|
A & = & \frac{ 5 }{ -45 } + \frac{ 5 }{ -45 } \\
|
||
|
|
A & = & \frac{ 5 + 5 }{ -45 } \\
|
||
|
|
A & = & \frac{ 10 }{ -45 } \\
|
||
|
|
A & = & \frac{ -10 }{ 45 } \\
|
||
|
|
A & = & \frac{ -2 \times 5 }{ 9 \times 5 } \\
|
||
|
|
A & = & \frac{ -2 }{ 9 }
|
||
|
|
\end{eqnarray*}
|
||
|
|
\end{solution}
|
||
|
|
|
||
|
|
|
||
|
|
\subpart $\displaystyle C = \frac{ -3 }{ 3 } + \frac{ 10 }{ 2 }$
|
||
|
|
\begin{solution}
|
||
|
|
|
||
|
|
\begin{eqnarray*}
|
||
|
|
A & = & \frac{ -3 }{ 3 } + \frac{ 10 }{ 2 } \\
|
||
|
|
A & = & \frac{ -3 \times 2 }{ 3 \times 2 } + \frac{ 10 \times 3 }{ 2 \times 3 } \\
|
||
|
|
A & = & \frac{ -6 }{ 6 } + \frac{ 30 }{ 6 } \\
|
||
|
|
A & = & \frac{ -6 + 30 }{ 6 } \\
|
||
|
|
A & = & 4
|
||
|
|
\end{eqnarray*}
|
||
|
|
\end{solution}
|
||
|
|
\subpart $\displaystyle D = \frac{ -6 }{ -8 } + 3$
|
||
|
|
\begin{solution}
|
||
|
|
|
||
|
|
\begin{eqnarray*}
|
||
|
|
A & = & \frac{ -6 }{ -8 } + 3 \\
|
||
|
|
A & = & \frac{ -6 \times 1 }{ -8 \times 1 } + \frac{ 3 \times ( -8 ) }{ 1 \times ( -8 ) } \\
|
||
|
|
A & = & \frac{ -6 }{ -8 } + \frac{ -24 }{ -8 } \\
|
||
|
|
A & = & \frac{ -6 - 24 }{ -8 } \\
|
||
|
|
A & = & \frac{ -30 }{ -8 } \\
|
||
|
|
A & = & \frac{ 30 }{ 8 } \\
|
||
|
|
A & = & \frac{ 15 \times 2 }{ 4 \times 2 } \\
|
||
|
|
A & = & \frac{ 15 }{ 4 }
|
||
|
|
\end{eqnarray*}
|
||
|
|
\end{solution}
|
||
|
|
\end{subparts}
|
||
|
|
\end{multicols}
|
||
|
|
|
||
|
|
\part Mettre les expressions suivantes sur le même dénominateur
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
\begin{multicols}{2}
|
||
|
|
\begin{subparts}
|
||
|
|
\subpart $\displaystyle A = \frac{ 8 }{ 4 x } \times ( -10 )$
|
||
|
|
\subpart $\displaystyle B = \frac{ 4 }{ -4 } + \frac{ -4 x }{ 36 }$
|
||
|
|
\subpart $\displaystyle C = \frac{ -7 x }{ 3 } + \frac{ 3 }{ -4 x }$
|
||
|
|
\subpart $\displaystyle D = \frac{ -7 }{ -5 x } + 4$
|
||
|
|
\end{subparts}
|
||
|
|
\end{multicols}
|
||
|
|
\end{parts}
|
||
|
|
|
||
|
|
\end{questions}
|
||
|
|
|
||
|
|
\end{document}
|
||
|
|
|
||
|
|
%%% Local Variables:
|
||
|
|
%%% mode: latex
|
||
|
|
%%% TeX-master: "master"
|
||
|
|
%%% End:
|