142 lines
5.2 KiB
TeX
142 lines
5.2 KiB
TeX
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\documentclass[a5paper,10pt]{article}
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\usepackage{myXsim}
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\usepackage{tasks}
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% Title Page
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\title{DM1 \hfill MAGRO Robin}
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\tribe{TST}
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\date{Toussain 2020}
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\begin{document}
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\maketitle
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\begin{exercise}[subtitle={Fractions}]
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Faire les calculs avec les fraction suivants
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\begin{multicols}{3}
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\begin{enumerate}
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\item $A = \dfrac{- 2}{6} - \dfrac{- 4}{6}$
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\item $B = \dfrac{5}{6} - \dfrac{- 2}{48}$
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\item $C = \dfrac{6}{9} + \dfrac{8}{8}$
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\item $D = \dfrac{- 5}{3} - 10$
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\item $E = \dfrac{- 9}{7} \times \dfrac{- 9}{6}$
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\item $F = \dfrac{- 3}{6} \times - 1$
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\end{enumerate}
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\end{multicols}
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\end{exercise}
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\begin{solution}
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\begin{enumerate}
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\item
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\[
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\dfrac{- 2}{6} - \dfrac{- 4}{6}=\dfrac{- 2}{6} + \dfrac{4}{6}=\dfrac{- 2 + 4}{6}=\dfrac{2}{6}
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\]
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\item
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\[
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\dfrac{5}{6} - \dfrac{- 2}{48}=\dfrac{5}{6} + \dfrac{2}{48}=\dfrac{5 \times 8}{6 \times 8} + \dfrac{2}{48}=\dfrac{40}{48} + \dfrac{2}{48}=\dfrac{40 + 2}{48}=\dfrac{42}{48}
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\]
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\item
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\[
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\dfrac{6}{9} + \dfrac{8}{8}=\dfrac{6 \times 8}{9 \times 8} + \dfrac{8 \times 9}{8 \times 9}=\dfrac{48}{72} + \dfrac{72}{72}=\dfrac{48 + 72}{72}=\dfrac{120}{72}
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\]
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\item
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\[
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\dfrac{- 5}{3} - 10=\dfrac{- 5}{3} + \dfrac{- 10}{1}=\dfrac{- 5}{3} + \dfrac{- 10 \times 3}{1 \times 3}=\dfrac{- 5}{3} + \dfrac{- 30}{3}=\dfrac{- 5 - 30}{3}=\dfrac{- 35}{3}
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\]
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\item
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\[
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\dfrac{- 9}{7} \times \dfrac{- 9}{6}=\dfrac{- 9 \times - 9}{7 \times 6}=\dfrac{81}{42}
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\]
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\item
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\[
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\dfrac{- 3}{6} \times - 1=\dfrac{- 3 \times - 1}{6}=\dfrac{3}{6}
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\]
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\end{enumerate}
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\end{solution}
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\begin{exercise}[subtitle={Développer réduire}]
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Développer puis réduire les expressions suivantes
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\begin{multicols}{2}
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\begin{enumerate}
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\item $A = (3x + 4)(- 1x + 4)$
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\item $B = (3x - 2)(- 3x - 2)$
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\item $C = (- 4x - 2)^{2}$
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\item $D = 5 + x(2x + 7)$
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\item $E = - 8x^{2} + x(6x + 2)$
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\item $F = - 2(x - 10)(x - 3)$
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\end{enumerate}
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\end{multicols}
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\end{exercise}
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\begin{solution}
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\begin{enumerate}
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\item
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\begin{align*}
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A &= (3x + 4)(- 1x + 4)\\&= 3x \times - x + 3x \times 4 + 4 \times - x + 4 \times 4\\&= 3 \times - 1 \times x^{1 + 1} + 4 \times 3 \times x + 4 \times - 1 \times x + 16\\&= 12x - 4x - 3x^{2} + 16\\&= (12 - 4) \times x - 3x^{2} + 16\\&= - 3x^{2} + 8x + 16
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\end{align*}
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\item
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\begin{align*}
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B &= (3x - 2)(- 3x - 2)\\&= 3x \times - 3x + 3x \times - 2 - 2 \times - 3x - 2 \times - 2\\&= 3 \times - 3 \times x^{1 + 1} - 2 \times 3 \times x - 2 \times - 3 \times x + 4\\&= - 6x + 6x - 9x^{2} + 4\\&= (- 6 + 6) \times x - 9x^{2} + 4\\&= 0x - 9x^{2} + 4\\&= - 9x^{2} + 4
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\end{align*}
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\item
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\begin{align*}
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C &= (- 4x - 2)^{2}\\&= (- 4x - 2)(- 4x - 2)\\&= - 4x \times - 4x - 4x \times - 2 - 2 \times - 4x - 2 \times - 2\\&= - 4 \times - 4 \times x^{1 + 1} - 2 \times - 4 \times x - 2 \times - 4 \times x + 4\\&= 8x + 8x + 16x^{2} + 4\\&= (8 + 8) \times x + 16x^{2} + 4\\&= 16x^{2} + 16x + 4
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\end{align*}
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\item
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\begin{align*}
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D &= 5 + x(2x + 7)\\&= 5 + x \times 2x + x \times 7\\&= 2x^{2} + 7x + 5
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\end{align*}
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\item
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\begin{align*}
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E &= - 8x^{2} + x(6x + 2)\\&= - 8x^{2} + x \times 6x + x \times 2\\&= - 8x^{2} + 6x^{2} + 2x\\&= - 8x^{2} + 6x^{2} + 2x\\&= (- 8 + 6) \times x^{2} + 2x\\&= - 2x^{2} + 2x
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\end{align*}
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\item
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\begin{align*}
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F &= - 2(x - 10)(x - 3)\\&= (- 2x - 2 \times - 10)(x - 3)\\&= (- 2x + 20)(x - 3)\\&= - 2x \times x - 2x \times - 3 + 20x + 20 \times - 3\\&= - 3 \times - 2 \times x - 60 - 2x^{2} + 20x\\&= 6x - 60 - 2x^{2} + 20x\\&= - 2x^{2} + 6x + 20x - 60\\&= - 2x^{2} + (6 + 20) \times x - 60\\&= - 2x^{2} + 26x - 60
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\end{align*}
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\end{enumerate}
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\end{solution}
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\begin{exercise}[subtitle={Étude de fonctions}]
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Soit $f(x) = 6x^{2} - 150$ une fonction définie sur $\R$.
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\begin{enumerate}
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\item Calculer les valeurs suivantes
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\[
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f(1) \qquad f(-2)
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\]
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\item Dériver la fonction $f$
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\item Étudier le signe de $f'$ puis en déduire les variations de $f$.
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\item Est-ce que $f$ admet un maximum? un minimum? Calculer sa valeur.
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\end{enumerate}
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\end{exercise}
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\begin{solution}
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\begin{enumerate}
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\item On remplace $x$ par les valeurs demandées
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\[
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f(1) = 6 \times 1^{2} - 150=6 \times 1 - 150=6 - 150=- 144
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\]
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\[
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f(-1) = 6 \times - 1^{2} - 150=6 \times 1 - 150=6 - 150=- 144
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\]
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\item Pas de solutions automatiques.
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\item Pas de solutions automatiques.
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\end{enumerate}
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\end{solution}
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%\printsolutionstype{exercise}
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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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