210 lines
7.0 KiB
TeX
210 lines
7.0 KiB
TeX
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\documentclass[a4paper,12pt, table]{/media/documents/Cours/Prof/Enseignements/2014-2015/tools/style/classDS}
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\usepackage{/media/documents/Cours/Prof/Enseignements/2014-2015/2014_2015}
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%\geometry{left=10mm,right=10mm, top=10mm, bottom=10mm}
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% Title Page
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\titre{4}
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% \seconde \premiereS \PSTMG \TSTMG
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\classe{\seconde}
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\date{6 mai 2015}
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%\duree{1 heure}
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\sujet{10}
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% DS DSCorr DM DMCorr Corr
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\typedoc{DM}
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%\printanswers
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\begin{document}
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\maketitle
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Vous devez rendre le sujet avec la copie.
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\begin{questions}
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\question
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\begin{parts}
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\part Développer et simplifier les expressions suivantes
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\begin{subparts}
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\begin{multicols}{2}
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\subpart $A = ( -4 x - 8 ) ( -4 - ( -5 x ) )$
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\begin{solution}
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\begin{eqnarray*}
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A & = & ( -4 x - 8 ) ( -4 - ( -5 x ) ) \\
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A & = & ( - 4 x - 8 ) ( -4 - ( - 5 x ) ) \\
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A & = & ( - 4 x - 8 ) ( - 4 - ( - 5 x ) ) \\
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A & = & - 4 x - 8 ( - 4 + 5 x ) \\
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A & = & - 20 x^{ 2 } - 24 x + 32
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\end{eqnarray*}
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\end{solution}
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\subpart $B = ( -9 x - 8 )^{ 2 } - 7$
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\begin{solution}
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\begin{eqnarray*}
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A & = & ( -9 x - 8 )^{ 2 } - 7 \\
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A & = & ( - 9 x - 8 )^{ 2 } - 7 \\
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A & = & ( - 9 x - 8 )^{ 2 } - 7 \\
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A & = & - 9 x - 8^{ 2 } - 7 \\
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A & = & - 9 x - 8 ( - 9 x - 8 ) - 7 \\
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A & = & -9 \times ( -9 ) x^{ 2 } + ( -8 \times ( -9 ) - 9 \times ( -8 ) ) x - 8 \times ( -8 ) - 7 \\
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A & = & 81 x^{ 2 } + ( 72 + 72 ) x + 64 - 7 \\
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A & = & 81 x^{ 2 } + 144 x + 57
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\end{eqnarray*}
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\end{solution}
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\subpart $C = -7 x - 3 + 4 ( -5 x - 1 )^{ 2 }$
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\begin{solution}
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\begin{eqnarray*}
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A & = & -7 x - 3 + 4 ( -5 x - 1 )^{ 2 } \\
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A & = & - 7 x - 3 + 4 ( - 5 x - 1 )^{ 2 } \\
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A & = & - 7 x - 3 + 4 ( - 5 x - 1 )^{ 2 } \\
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A & = & - 7 x - 3 + 4 - 5 x - 1^{ 2 } \\
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A & = & - 7 x - 3 + 4 - 5 x - 1^{ 2 } \\
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A & = & - 7 x - 3 + 4 - 5 x - 1 ( - 5 x - 1 ) \\
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A & = & - 7 x - 3 + 4 ( -5 \times ( -5 ) x^{ 2 } + ( -1 \times ( -5 ) - 5 \times ( -1 ) ) x - 1 \times ( -1 ) ) \\
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A & = & - 7 x - 3 + 4 ( 25 x^{ 2 } + ( 5 + 5 ) x + 1 ) \\
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A & = & - 7 x - 3 + 4 ( 25 x^{ 2 } + 10 x + 1 ) \\
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A & = & - 7 x - 3 + 4 ( 25 x^{ 2 } + 10 x + 1 ) \\
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A & = & - 7 x - 3 + 4 \times 25 x^{ 2 } + 4 \times 10 x + 4 \\
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A & = & 100 x^{ 2 } + 33 x + 1
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\end{eqnarray*}
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\end{solution}
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\end{multicols}
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\end{subparts}
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\part Factoriser les expressions suivantes
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\begin{subparts}
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\begin{multicols}{2}
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\subpart $A = 1 x^{ 2 } - x$
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\subpart $B = 100 x^{ 2 } + 9 + 60 x$
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\subpart $C = 64 x^{ 2 } - 49$
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\subpart $D = 100 x^{ 2 } - 120 x + 36$
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\end{multicols}
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\end{subparts}
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\part Résoudre les équations suivantes
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\begin{subparts}
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\begin{multicols}{2}
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\subpart $2 x + 1 = 0$
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\subpart $- x - 10 = 6 x - 8$
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\columnbreak
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\subpart $5 x^{ 2 } + 2 x - 10 = 5x^2$
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\subpart $( -9 x + 3 ) ( 6 x - 10 ) = 0$
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\end{multicols}
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\end{subparts}
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\end{parts}
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\question
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\begin{parts}
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\part Soit $A(-10 ; -3)$, $B(2 ; 3)$, $C(-3 ; -10)$ et $D(7 ; -5)$. Est-ce que les vecteurs $\vec{AB}$ et $\vec{CD}$ sont colinéaires?
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\part Soit $A(-7 ; -6)$, $B(10 ; 1)$, $C(-6 ; -2)$ et $D(-10 ; -4)$. Est-ce que les vecteurs $\vec{AB}$ et $\vec{CD}$ sont colinéaires?
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\part Soit $A(-4 ; 4)$, $B(-4 ; 4)$, $C(-4 ; 6)$ et $D(10 ; -9)$. Est-ce que les droites $(AC)$ et $(BD)$ sont colinéaires?
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\end{parts}
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\question
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\begin{parts}
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\part Faire les calculs suivants en \textbf{détaillant les étapes} et en simplifiant les fractions.
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\begin{multicols}{2}
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\begin{subparts}
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\subpart $\displaystyle A = \frac{ 1 }{ -9 } \times 7$
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\begin{solution}
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\begin{eqnarray*}
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A & = & \frac{ 1 }{ -9 } \times 7 \\
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A & = & \frac{ 1 \times 7 }{ -9 } \\
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A & = & \frac{ 7 }{ -9 } \\
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A & = & \frac{ -7 }{ 9 }
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\end{eqnarray*}
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\end{solution}
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\subpart $\displaystyle B = \frac{ -9 }{ 3 } + \frac{ 3 }{ -15 }$
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\begin{solution}
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\begin{eqnarray*}
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A & = & \frac{ -9 }{ 3 } + \frac{ 3 }{ -15 } \\
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A & = & \frac{ -9 \times ( -5 ) }{ 3 \times ( -5 ) } + \frac{ 3 \times 1 }{ -15 \times 1 } \\
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A & = & \frac{ 45 }{ -15 } + \frac{ 3 }{ -15 } \\
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A & = & \frac{ 45 + 3 }{ -15 } \\
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A & = & \frac{ 48 }{ -15 } \\
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A & = & \frac{ -48 }{ 15 } \\
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A & = & \frac{ -16 \times 3 }{ 5 \times 3 } \\
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A & = & \frac{ -16 }{ 5 }
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\end{eqnarray*}
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\end{solution}
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\subpart $\displaystyle C = \frac{ 2 }{ -7 } + \frac{ -1 }{ 4 }$
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\begin{solution}
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\begin{eqnarray*}
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A & = & \frac{ 2 }{ -7 } + \frac{ -1 }{ 4 } \\
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A & = & \frac{ 2 \times 4 }{ -7 \times 4 } + \frac{ -1 \times ( -7 ) }{ 4 \times ( -7 ) } \\
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A & = & \frac{ 8 }{ -28 } + \frac{ 7 }{ -28 } \\
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A & = & \frac{ 8 + 7 }{ -28 } \\
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A & = & \frac{ 15 }{ -28 } \\
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A & = & \frac{ -15 }{ 28 }
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\end{eqnarray*}
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\end{solution}
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\subpart $\displaystyle D = \frac{ 2 }{ 4 } + 8$
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\begin{solution}
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\begin{eqnarray*}
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A & = & \frac{ 2 }{ 4 } + 8 \\
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A & = & \frac{ 2 \times 1 }{ 4 \times 1 } + \frac{ 8 \times 4 }{ 1 \times 4 } \\
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A & = & \frac{ 2 }{ 4 } + \frac{ 32 }{ 4 } \\
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A & = & \frac{ 2 + 32 }{ 4 } \\
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A & = & \frac{ 34 }{ 4 } \\
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A & = & \frac{ 17 \times 2 }{ 2 \times 2 } \\
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A & = & \frac{ 17 }{ 2 }
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\end{eqnarray*}
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\end{solution}
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\end{subparts}
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\end{multicols}
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\part Mettre les expressions suivantes sur le même dénominateur
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\begin{multicols}{2}
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\begin{subparts}
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\subpart $\displaystyle A = \frac{ 9 }{ 4 x } \times 10$
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\subpart $\displaystyle B = \frac{ -9 }{ -7 } + \frac{ 7 x }{ 28 }$
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\subpart $\displaystyle C = \frac{ 6 x }{ 4 } + \frac{ -10 }{ -3 x }$
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\subpart $\displaystyle D = \frac{ -3 }{ -4 x } - 1$
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\end{subparts}
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\end{multicols}
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\end{parts}
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\end{questions}
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\end{document}
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: "master"
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%%% End:
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