Import work from year 2013-2014
This commit is contained in:
Benjamin Bertrand
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\documentclass{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classConn}
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% Title Page
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\title{}
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\author{}
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\date{}
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\begin{document}
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\begin{multicols}{2}
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Nom - Prénom:
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\section{Connaissance}
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\begin{enumerate}
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\item Donner les deux identités remarquables
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~\\[0.5cm]
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.\dotfill
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~\\[0.5cm]
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.\dotfill
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~\\[0.5cm]
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\item Mettre sous la forme $a^2$ en précisant la valeur de $a$
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\begin{eqnarray*}
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A = 25x^2 = \dots \hspace{2cm} a = \dots\\
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~\\[0.5cm]
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B = 64 = \dots \hspace{2cm} a = \dots
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\end{eqnarray*}
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\item Mettre sous la forme $2ab$ en précisant la valeur de $a$ et de $b$
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\begin{eqnarray*}
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C = 30x = \dots \hspace{2cm} a = \dots \hspace{1cm} b = \dots\\
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\end{eqnarray*}
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\item Donner la définition d'une \textbf{issue} d'une experience aléatoire:
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~\\[0.5cm]
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.\dotfill
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~\\[0.5cm]
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.\dotfill
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~\\[0.5cm]
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\end{enumerate}
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\columnbreak
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Nom - Prénom
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\section{Connaissance}
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\begin{enumerate}
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\item Donner les deux identités remarquables
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~\\[0.5cm]
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.\dotfill
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~\\[0.5cm]
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.\dotfill
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~\\[0.5cm]
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\item Mettre sous la forme $a^2$ en précisant la valeur de $a$
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\begin{eqnarray*}
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A = 49x^2 = \dots \hspace{2cm} a = \dots\\
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~\\[0.5cm]
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B = 100 = \dots \hspace{2cm} a = \dots
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\end{eqnarray*}
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\item Mettre sous la forme $2ab$ en précisant la valeur de $a$ et de $b$
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\begin{eqnarray*}
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C = 12x = \dots \hspace{2cm} a = \dots \hspace{1cm} b = \dots\\
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\end{eqnarray*}
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\item Donner la définition d'un \textbf{évènement} d'une experience aléatoire:
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~\\[0.5cm]
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.\dotfill
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~\\[0.5cm]
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.\dotfill
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~\\[0.5cm]
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\end{enumerate}
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\end{multicols}
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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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\documentclass[a4paper,10pt]{beamer}
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\usepackage[utf8]{inputenc}
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\usepackage[french]{babel}
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\usepackage{graphicx}
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\usepackage{thumbpdf}
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\usepackage{wasysym}
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\usepackage{ucs}
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\usepackage{pgf,pgfarrows,pgfnodes,pgfautomata,pgfheaps,pgfshade}
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\usepackage{verbatim}
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\usepackage{subfig}
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\usepackage{amssymb}
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\begin{document}
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\begin{frame}{Solution l'exercice 1}
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\begin{itemize}
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\item La solution de l'équation $2x + 1 = 0$ est $\mathbf{x = \frac{-1}{2}}$
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\vfill
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\item La solution de l'équation $6x + 12 = 0$ est $\mathbf{x = -2}$
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\vfill
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\item La solution de l'équation $3x - 3 = 0$ est $\mathbf{x = 1}$
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\vfill
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\item La solution de l'équation $8x - 4 = 0$ est $\mathbf{x = \frac{1}{2}}$
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\vfill
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\item La solution de l'équation $-6x - 3 = 0$ est $\mathbf{x = \frac{-1}{2}}$
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\vfill
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\item La solution de l'équation $9 + 3x = 0$ est $\mathbf{x = -3}$
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\vfill
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\item La solution de l'équation $5 + 3x = 0$ est $\mathbf{x = \frac{-5}{3}}$
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\vfill
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\item La solution de l'équation $\frac{2}{3}x + 3 = 0$ est $\mathbf{x = \frac{-9}{2}}$
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\end{itemize}
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\end{frame}
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\end{document}
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\documentclass[a4paper,10pt]{beamer}
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\usepackage[utf8]{inputenc}
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\usepackage[french]{babel}
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\usepackage{graphicx}
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\usepackage{thumbpdf}
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\usepackage{wasysym}
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\usepackage{ucs}
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\usepackage{pgf,pgfarrows,pgfnodes,pgfautomata,pgfheaps,pgfshade}
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\usepackage{verbatim}
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\usepackage{subfig}
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\usepackage{amssymb}
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\usepackage{multicol}
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\begin{document}
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\begin{frame}{Solutions de l'exercice 2}
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\begin{enumerate}
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\item Les solutions sont $x = \frac{-3}{2}$ et $x = \frac{-5}{4}$.
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\vfill
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\item Les solutions sont $x = \frac{-13}{6}$ et $x = \frac{-44}{5}$.
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\vfill
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\item Les solutions sont $x = \frac{1}{3}$ et $x = 9$.
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\vfill
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\item Les solutions sont $x = 3$ et $x = \frac{-1}{2}$.
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\vfill
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\item Les solutions sont $x = 1$ et $x = 2$.
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\vfill
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\item La solution est $x = \frac{-3}{2}$.
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\vfill
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\item Les solutions sont $x = \frac{-8}{3}$ et $x = \frac{8}{3}$.
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\vfill
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\item La solution est $x = \frac{1}{7}$.
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\end{enumerate}
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\end{frame}
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\end{document}
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Notes sur correction
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####################
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:date: 2014-07-01
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:modified: 2014-07-01
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:tags: Nombres Calculs
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:category: 3e
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:authors: Benjamin Bertrand
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:summary: Pas de résumé, note créée automatiquement parce que je ne l'avais pas bien fait...
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`Lien vers corr_eq_1.tex <corr_eq_1.tex>`_
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`Lien vers corr_eq_1.pdf <corr_eq_1.pdf>`_
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`Lien vers corr_eq_2.tex <corr_eq_2.tex>`_
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`Lien vers corr_eq_2.pdf <corr_eq_2.pdf>`_
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\documentclass[a4paper,12pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
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% Title Page
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\title{Identités remarquables - Exercices}
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\author{}
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\date{}
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\fancyhead[L]{Troisième}
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\fancyhead[C]{\Thetitle}
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\fancyhead[R]{\thepage}
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\begin{document}
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\thispagestyle{fancy}
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\begin{Exo}
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\begin{enumerate}
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\item Relier les expressions égales entres elles.
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\begin{minipage}[c]{0.2\textwidth}
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\flushright
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$4x^2 + 4x + 1 \qquad \bullet$ \\[0.5cm]
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$64x^2 - 48x + 9 \qquad \bullet$ \\[0.5cm]
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$36x^2 + 60x + 25 \qquad \bullet$ \\[0.5cm]
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$36x^2 - 60x + 25 \qquad \bullet$
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\end{minipage}
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\hspace{2cm}
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\begin{minipage}[c]{0.1\textwidth}
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\begin{itemize}
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\item $(8x - 3)^2$
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\item $(6x + 5)^2$
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\item $(2x + 1)^2$
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\item $(6x - 5)^2$
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\item $(36x + 25)^2$
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\item $(4x + 1)^2$
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\item $(2x - 1)^2$
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\item $(8x + 3)^2$
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\end{itemize}
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\end{minipage}
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\item Factoriser l'expression suivante
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\begin{eqnarray*}
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A & = & 25x^2 + 30x + 9
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\end{eqnarray*}
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\end{enumerate}
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\end{Exo}
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\vspace{1cm}
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\begin{Exo}
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\begin{enumerate}
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\item Relier les expressions égales entres elles.
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\begin{minipage}[c]{0.2\textwidth}
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\flushright
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$4x + 4x^2 + 1 \qquad \bullet$ \\[0.5cm]
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$9 - 48x + 64x^2 \qquad \bullet$ \\[0.5cm]
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$4 + 49x^2 - 28x \qquad \bullet$ \\[0.5cm]
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$16x + 16x^2 + 4 \qquad \bullet$
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\end{minipage}
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\hspace{2cm}
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\begin{minipage}[c]{0.1\textwidth}
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\begin{itemize}
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\item $(2x + 1)^2$
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\item $(8x - 3)^2$
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\item $(7x + 3)^2$
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\item $(2x + 4)^2$
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\item $(2x - 1)^2$
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\item $(3 - 7x)^2$
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\item $(2 + 4x)^2$
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\item $(8x + 3)^2$
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\end{itemize}
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\end{minipage}
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\item Factoriser les expressions suivantes
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\begin{eqnarray*}
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A & = & 4 + 25x^2 + 20x\\
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B & = & -72x + 81x^2 + 16
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\end{eqnarray*}
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\end{enumerate}
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\end{Exo}
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\begin{Exo}
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\begin{enumerate}
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\item Relier les expressions égales entres elles.
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\begin{minipage}[c]{0.2\textwidth}
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\flushright
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$x^2 + 2x + 1 \qquad \bullet$ \\[0.5cm]
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$2x^2 + 6\sqrt{2}x + 9 \qquad \bullet$ \\[0.5cm]
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$3x^2 + 4\sqrt{3}x + 4 \qquad \bullet$ \\[0.5cm]
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$9x^2 + 6\sqrt{2}x + 2 \qquad \bullet$
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\end{minipage}
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\hspace{2cm}
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\begin{minipage}[c]{0.1\textwidth}
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\begin{itemize}
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\item $(x + 1)^2$
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\item $(\sqrt{2}x - 3)^2$
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\item $(\sqrt{2}x + 3)^2$
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\item $(3x - \sqrt{2})^2$
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\item $(x - 1)^2$
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\item $(\sqrt{3}x + 2)^2$
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\item $(3x + \sqrt{2})^2$
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\item $(\sqrt{3}x - 2)^2$
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\end{itemize}
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\end{minipage}
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\item Factoriser les expressions suivantes
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\begin{eqnarray*}
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A & = & 2x^2 + 8\sqrt{2}x + 16 \\
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B & = & 3x^2 + 10\sqrt{3}x + 25 \\
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C & = & 3x^2 + 2\sqrt{6}x + 2 \\
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\end{eqnarray*}
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\end{enumerate}
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\end{Exo}
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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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\documentclass[a4paper,12pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
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% Title Page
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\title{Identités remarquables - Exercices}
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\author{}
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\date{}
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\fancyhead[L]{Troisième}
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\fancyhead[C]{\Thetitle}
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\fancyhead[R]{\thepage}
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\begin{document}
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\thispagestyle{fancy}
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\begin{Exo}
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Rappeler les deux identités remarquables.
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\begin{eqnarray*}
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(a + b)^2 & = & \hspace{3cm} \\[0.5cm]
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(a - b)^2 & = & \hspace{3cm}
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\end{eqnarray*}
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Développer puis réduire pour découvrir la 3e identité remarquable.
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\begin{eqnarray*}
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(a + b)(a - b) & = & \parbox{1cm}{\dotfill} \times \parbox{1cm}{\dotfill} + \parbox{1cm}{\dotfill} \times \parbox{1cm}{\dotfill} + \parbox{1cm}{\dotfill} \times \parbox{1cm}{\dotfill} + \parbox{1cm}{\dotfill} \times \parbox{1cm}{\dotfill} \\[0.5cm]
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& = & \parbox{1cm}{\dotfill} + \parbox{1cm}{\dotfill} + \parbox{1cm}{\dotfill} + \parbox{1cm}{\dotfill} \\[0.5cm]
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& = & \parbox{1cm}{\dotfill} - \parbox{1cm}{\dotfill}
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\end{eqnarray*}
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On en déduit la troisième identité remarquable
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\begin{center}
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\framebox{\parbox{0.2\textwidth}{
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\begin{eqnarray*}
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(a + b)(a - b) & = & \parbox{1cm}{\dotfill} - \parbox{1cm}{\dotfill}
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\end{eqnarray*}
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}}
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\end{center}
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\end{Exo}
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\begin{Exo}
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\begin{enumerate}
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\item Relier les expressions égales entres elles (en utilisant la 3e identité remarquable).
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\begin{minipage}[c]{0.2\textwidth}
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\flushright
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$(2x + 1)(2x - 1) \qquad \bullet$ \\[0.5cm]
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$(4x + 2)(4x - 2) \qquad \bullet$ \\[0.5cm]
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$(8x - 1)(8x + 1) \qquad \bullet$ \\[0.5cm]
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$(5 + 2x)(5 - 2x) \qquad \bullet$
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\end{minipage}
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\hspace{1cm}
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\begin{minipage}[c]{0.2\textwidth}
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\begin{itemize}
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\item $4x^2 + 16x - 2$
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\item $4x^2 - 1$
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\item $4x - 1$
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\item $25 - 4x$
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\item $16x^2 - 4$
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\item $64x - 1$
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\item $4x^2 + 1$
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\item $4x - 25$
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\end{itemize}
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\end{minipage}
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\item Développer les expressions suivantes
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\begin{eqnarray*}
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A = (3x + 5)(3x - 5) \hspace{2cm} B = (7x - 4)(7x + 4)
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\end{eqnarray*}
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\end{enumerate}
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\end{Exo}
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\vspace{1cm}
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\begin{Exo}
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\begin{enumerate}
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\item Relier les expressions égales entres elles.
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\begin{minipage}[c]{0.15\textwidth}
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\flushright
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$4x^2 - 9 \qquad \bullet$ \\[0.5cm]
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$64x^2 - 16 \qquad \bullet$ \\[0.5cm]
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$49x^2 - 81\qquad \bullet$ \\[0.5cm]
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$36 - 9x^2 \qquad \bullet$
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\end{minipage}
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\hspace{2cm}
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\begin{minipage}[c]{0.2\textwidth}
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\begin{itemize}
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\item $(4x - 9)^2$
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\item $(3x + 6)(3x - 6)$
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\item $(7x + 9)(9 - 7x)$
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\item $(8x + 4)^2$
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\item $(4x + 9)(4x - 9)$
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\item $(7x + 9)(7x - 9)$
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\item $(8x - 4)(8x + 4)$
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\item $(6 - 3x)(6 + 3x)$
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\end{itemize}
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\end{minipage}
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\item Factoriser les expressions suivantes
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\begin{eqnarray*}
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A = 2x^2 - 9 \hspace{2cm} B = 9x^2 - 25 \\[0.5cm]
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C = 64x^2 - 1 \hspace{2cm} D = x^2 - 16
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\end{eqnarray*}
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\end{enumerate}
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\end{Exo}
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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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\documentclass[a4paper,12pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
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% Title Page
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\title{Identités remarquables - Exercices}
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\author{}
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\date{}
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\fancyhead[L]{Troisième}
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\fancyhead[C]{\Thetitle}
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\fancyhead[R]{\thepage}
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||||
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\begin{document}
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\thispagestyle{fancy}
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\begin{Exo}
|
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Rappeler les trois identités remarquables.
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\begin{center}
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||||
\framebox{\parbox{0.4\textwidth}{
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\begin{eqnarray*}
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(a + b)^2 & = & \hspace{5cm} \\[0.5cm]
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(a - b)^2 & = & \hspace{5cm} \\[0.5cm]
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(a + b)(a - c) & = & \hspace{5cm}
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\end{eqnarray*}
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}}
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\end{center}
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\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
\begin{enumerate}
|
||||
\item Relier les expressions égales entres elles (il faudra utiliser les 3 identités remarquables).
|
||||
|
||||
\begin{minipage}[c]{0.2\textwidth}
|
||||
\flushright
|
||||
$(2x + 1)(2x - 1) \qquad \bullet$ \\[0.5cm]
|
||||
$(4x + 2)^2 \qquad \bullet$ \\[0.5cm]
|
||||
$(8x - 1)(8x + 1) \qquad \bullet$ \\[0.5cm]
|
||||
$(5 - 2x)^2 \qquad \bullet$
|
||||
|
||||
\end{minipage}
|
||||
\hspace{1cm}
|
||||
\begin{minipage}[c]{0.2\textwidth}
|
||||
\begin{itemize}
|
||||
\item $25 + 20x + 4x^2$
|
||||
\item $16x^2 + 16x + 4$
|
||||
\item $4x^2 - 1$
|
||||
\item $16x^2 - 16x + 4$
|
||||
\item $64x^2 - 1$
|
||||
\item $64x^2 + 16x + 1$
|
||||
\item $4x^2 + 1$
|
||||
\item $25 - 20x + 4x^2$
|
||||
\end{itemize}
|
||||
|
||||
\end{minipage}
|
||||
|
||||
\item Développer les expressions suivantes en utilisant les identités remarquables.
|
||||
\begin{eqnarray*}
|
||||
A = (2x + 3)(2x - 3) &=& \parbox{3cm}{\dotfill} \\[0.5cm]
|
||||
B = (7x - 4)^2 &=& \parbox{3cm}{\dotfill}\\[0.5cm]
|
||||
C = (8x + 4)^2 &=& \parbox{3cm}{\dotfill}\\[0.5cm]
|
||||
D = (6x - 1)(6x + 1)&=&\parbox{3cm}{\dotfill}
|
||||
\end{eqnarray*}
|
||||
\end{enumerate}
|
||||
\end{Exo}
|
||||
|
||||
\vspace{1cm}
|
||||
|
||||
\begin{Exo}
|
||||
\begin{enumerate}
|
||||
\item Relier les expressions égales entres elles.
|
||||
|
||||
\begin{minipage}[c]{0.15\textwidth}
|
||||
\flushright
|
||||
$16x^2 - 9 \qquad \bullet$ \\[0.5cm]
|
||||
$9x^2 - 24x + 16 \qquad \bullet$ \\[0.5cm]
|
||||
$49x^2 + 112x + 64\qquad \bullet$ \\[0.5cm]
|
||||
$1 - 25x^2 \qquad \bullet$
|
||||
\end{minipage}
|
||||
\hspace{2cm}
|
||||
\begin{minipage}[c]{0.2\textwidth}
|
||||
\begin{itemize}
|
||||
\item $(1 + 5x)(1 - 5x)$
|
||||
\item $(4x - 3)^2$
|
||||
\item $(4x - 3)(4x + 3)$
|
||||
\item $(3x + 4)^2$
|
||||
\item $(7x + 8)(7x - 8)$
|
||||
\item $(1 - 5x)^2$
|
||||
\item $(3x - 4)^2$
|
||||
\item $(7x + 8)^2$
|
||||
\end{itemize}
|
||||
|
||||
\end{minipage}
|
||||
|
||||
\item Factoriser les expressions suivantes
|
||||
\begin{eqnarray*}
|
||||
A = 9x^2 - 9 \hspace{2cm} B = 4x^2 + 12x + 9 \\[0.5cm]
|
||||
C = 100x^2 - 121 \hspace{2cm} D = 49x^2 - 84x + 36
|
||||
\end{eqnarray*}
|
||||
\end{enumerate}
|
||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
En utilisant la troisième identité remarquable et en suivante l'exemple, factoriser les expressions suivantes.
|
||||
\begin{eqnarray*}
|
||||
A = (x + 2)^2 - 81 = (x + 2)^2 - 9^2 = (x + 2 + 9)(x + 2 - 9) = (x + 11)(x - 7) \\[0.5cm]
|
||||
\end{eqnarray*}
|
||||
\begin{eqnarray*}
|
||||
B = (x + 1)^2 - 4 &\hspace{2cm}& C = (x + 3)^2 - 9 \\[0.5cm]
|
||||
D = (2x + 1)^2 - 25 &\hspace{2cm}& E = 36 - (4x + 1)^2 \\[0.5cm]
|
||||
F = (2x - 1)^2 - (3x + 4)^2 &\hspace{2cm}& D = (3x - 1)^2 - (x + 1)^2
|
||||
\end{eqnarray*}
|
||||
\end{Exo}
|
||||
|
||||
|
||||
|
||||
|
||||
\end{document}
|
||||
|
||||
%%% Local Variables:
|
||||
%%% mode: latex
|
||||
%%% TeX-master: "master"
|
||||
%%% End:
|
||||
|
||||
Binary file not shown.
@@ -0,0 +1,122 @@
|
||||
\documentclass[a4paper,12pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
|
||||
|
||||
|
||||
% Title Page
|
||||
\title{Identités remarquables et équations- Exercices}
|
||||
\author{}
|
||||
\date{}
|
||||
|
||||
\fancyhead[L]{Troisième}
|
||||
\fancyhead[C]{\Thetitle}
|
||||
\fancyhead[R]{\thepage}
|
||||
|
||||
|
||||
\begin{document}
|
||||
\thispagestyle{fancy}
|
||||
|
||||
\begin{Exo}
|
||||
Rappeler les trois identités remarquables.
|
||||
|
||||
\begin{center}
|
||||
\framebox{\parbox{0.4\textwidth}{
|
||||
\begin{eqnarray*}
|
||||
(a + b)^2 & = & \hspace{5cm} \\[0.5cm]
|
||||
(a - b)^2 & = & \hspace{5cm} \\[0.5cm]
|
||||
(a + b)(a - c) & = & \hspace{5cm}
|
||||
\end{eqnarray*}
|
||||
}}
|
||||
\end{center}
|
||||
|
||||
|
||||
|
||||
\end{Exo}
|
||||
|
||||
|
||||
\begin{Exo}
|
||||
\begin{enumerate}
|
||||
\item Relier les expressions égales entres elles.
|
||||
|
||||
\begin{minipage}[c]{0.15\textwidth}
|
||||
\flushright
|
||||
$49x^2 + 126x + 81\qquad \bullet$ \\[0.5cm]
|
||||
$25x^2 - 16 \qquad \bullet$ \\[0.5cm]
|
||||
$x^2 - 8x + 16 \qquad \bullet$ \\[0.5cm]
|
||||
$100 - 4x^2 \qquad \bullet$
|
||||
\end{minipage}
|
||||
\hspace{2cm}
|
||||
\begin{minipage}[c]{0.2\textwidth}
|
||||
\begin{itemize}
|
||||
\item $(x - 4)^2$
|
||||
\item $(5x + 4)(5x - 4)$
|
||||
\item $(7x - 9)^2$
|
||||
\item $(x - 4)(x + 4)$
|
||||
\item $(7x + 9)^2$
|
||||
\item $(10 - 2x)(10 + 2x)$
|
||||
\item $(5x + 4)^2$
|
||||
\item $(2x - 10)(10 + 2x)$
|
||||
\end{itemize}
|
||||
|
||||
\end{minipage}
|
||||
|
||||
\item Factoriser les expressions suivantes
|
||||
\begin{eqnarray*}
|
||||
A = x^2 - 9 \hspace{2cm} B = x^2 + 6x + 9 \\[0.5cm]
|
||||
C = 9 - 4x^2 \hspace{2cm} D = 49x^2 + 84x + 36
|
||||
\end{eqnarray*}
|
||||
\end{enumerate}
|
||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
En utilisant la troisième identité remarquable et en suivante l'exemple, factoriser les expressions suivantes.
|
||||
|
||||
\framebox{\parbox{0.45\textwidth}{
|
||||
\begin{eqnarray*}
|
||||
A = (x + 2)^2 - 81 = (x + 2)^2 - 9^2 = (x + 2 + 9)(x + 2 - 9) = (x + 11)(x - 7)
|
||||
\end{eqnarray*}
|
||||
}}
|
||||
\begin{eqnarray*}
|
||||
B = (2x + 5)^2 - 16 &\hspace{2cm}& C = (7x + 2)^2 - 2 \\[0.5cm]
|
||||
D = (-x - 1)^2 - 25 &\hspace{2cm}& E = (4x + 1)^2 - 49
|
||||
\end{eqnarray*}
|
||||
\begin{center}
|
||||
\framebox{\parbox{0.45\textwidth}{
|
||||
On rappelle que quand on enlève les parenthèses avec un signe "-" devant, il faut changer le signe de \textbf{TOUS} les éléments entre les parenthèses.
|
||||
\begin{eqnarray*}
|
||||
F = (x + 2)^2 - (3x + 4)^2 &=& \left[ x + 2 + 3x + 4 \right]\left[ x + 2 - \mathbf{(3x + 4)} \right] \\
|
||||
&=& \left[ x + 3x + 2 + 4 \right]\left[ x + 2 \mathbf{- 3x - 4} \right] \\
|
||||
&=& \left( 4x + 6 \right)\left( -2x - 2 \right)
|
||||
\end{eqnarray*}
|
||||
}}
|
||||
\end{center}
|
||||
\begin{eqnarray*}
|
||||
G = (2x + 1)^2 - (x + 2)^2 &\hspace{2cm}& H = (3x - 1)^2 - (5x + 2)^2 \\
|
||||
I = (4x + 3)^2 - (3x - 4)^2 &\hspace{2cm}& J = (3x - 1)^2 - (x - 2)^2
|
||||
\end{eqnarray*}
|
||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
Résoudre les équations suivantes
|
||||
|
||||
%\framebox{\parbox{0.45\textwidth}{
|
||||
% \begin{eqnarray*}
|
||||
% 2x = 5 \quad \mbox{ On divise par 2 }\quad \frac{2x}{2} = \frac{5}{2} \quad \mbox{ donc }\quad x = \frac{5}{2}
|
||||
% \end{eqnarray*}
|
||||
%}}
|
||||
\begin{eqnarray*}
|
||||
a) \qquad 3x = 4 \hspace{2cm} b) \qquad -5x = 7 \\
|
||||
c) \qquad x + 3 = 4 \hspace{2cm} d) \qquad 2 + x = 9 \\
|
||||
e) \qquad 3x + 2 = 3 \hspace{2cm} f) \qquad 3x - 2 = 5
|
||||
\end{eqnarray*}
|
||||
|
||||
\end{Exo}
|
||||
|
||||
|
||||
|
||||
|
||||
\end{document}
|
||||
|
||||
%%% Local Variables:
|
||||
%%% mode: latex
|
||||
%%% TeX-master: "master"
|
||||
%%% End:
|
||||
|
||||
Binary file not shown.
@@ -0,0 +1,139 @@
|
||||
\documentclass[a4paper,12pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
|
||||
|
||||
\usepackage{multicol}
|
||||
|
||||
% Title Page
|
||||
\title{Identités remarquables et équations- Exercices}
|
||||
\author{}
|
||||
\date{}
|
||||
|
||||
\fancyhead[L]{Troisième}
|
||||
\fancyhead[C]{\Thetitle}
|
||||
\fancyhead[R]{\thepage}
|
||||
|
||||
|
||||
\begin{document}
|
||||
\thispagestyle{empty}
|
||||
|
||||
\begin{Exo}
|
||||
\exo{Équations de degrés 1}
|
||||
|
||||
\begin{center}
|
||||
\framebox{\parbox{0.4\textwidth}{
|
||||
Résoudre l'équation $3x + 5 = 0$.
|
||||
\begin{eqnarray*}
|
||||
3x + 5 = 0 & \hspace{1cm} & \mbox{On ajoute l'opposé de 5} \\
|
||||
3x + 5 \mathbf{+ (-5)} = \mathbf{-5} && \\
|
||||
3x = -5 & \hspace{1cm} & \mbox{On multiplie par l'inverse de 3} \\
|
||||
\mathbf{\frac{1}{3} \times }3x = \mathbf{ \frac{1}{3} \times }(-5) && \\
|
||||
x = \frac{-5}{3}
|
||||
\end{eqnarray*}
|
||||
La solution est $x = \frac{-5}{3}$.
|
||||
}}
|
||||
\end{center}
|
||||
|
||||
\begin{enumerate}
|
||||
\item Résoudre l'équation $4x + 7 = 0$.
|
||||
\begin{eqnarray*}
|
||||
4x + 7 = 0 & \hspace{0.5cm} & \mbox{On ajoute l'opposé de \parbox{1cm}{\dotfill}} \\[0.5cm]
|
||||
4x + 7 + \parbox{1.5cm}{\dotfill}= \parbox{1.5cm}{\dotfill}&& \\[0.5cm]
|
||||
4x = \parbox{1cm}{\dotfill}& \hspace{0.5cm} & \mbox{On multiplie par l'inverse de \parbox{1cm}{\dotfill}} \\[0.5cm]
|
||||
\parbox{1.5cm}{\dotfill} \times 4x = \parbox{1.5cm}{\dotfill} \times \parbox{1cm}{\dotfill} && \\[0.5cm]
|
||||
x = \frac{\parbox{1cm}{\dotfill}}{\parbox{1cm}{\dotfill}}
|
||||
\end{eqnarray*}
|
||||
La solution est \parbox{2cm}{\dotfill}.
|
||||
|
||||
\item Résoudre les équations suivantes
|
||||
\begin{multicols}{2}
|
||||
\begin{enumerate}
|
||||
\item $2x + 1 = 0$
|
||||
\item $6x + 12 = 0$
|
||||
\item $3x - 3 = 0$
|
||||
\item $8x - 4 = 0$
|
||||
\columnbreak
|
||||
\item $-6x - 3 = 0$
|
||||
\item $9 + 3x = 0$
|
||||
\item $5 + 3x = 0$
|
||||
\item $\frac{2}{3}x + 3 = 0$
|
||||
\end{enumerate}
|
||||
\end{multicols}
|
||||
\end{enumerate}
|
||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
\exo{Équation produit}
|
||||
\begin{center}
|
||||
\framebox{\parbox{0.45\textwidth}{
|
||||
Résoudre l'équation $(3x + 5)(2x + 4) = 0$.
|
||||
|
||||
$(3x + 5)(2x + 4) = 0$ donc
|
||||
\begin{center}
|
||||
$3x + 5 = 0$ \hspace{1cm} soit \hspace{1cm} $2x + 4 = 0$
|
||||
\end{center}
|
||||
|
||||
\begin{multicols}{2}
|
||||
Soit
|
||||
\begin{eqnarray*}
|
||||
3x + 5 = 0 \\
|
||||
3x + 5 \mathbf{+ (-5)} = \mathbf{-5}\\
|
||||
3x = -5 \\
|
||||
\mathbf{\frac{1}{3} \times } 3x = \mathbf{ \frac{1}{3} \times } (-5) \\
|
||||
x = \frac{-5}{3}
|
||||
\end{eqnarray*}
|
||||
|
||||
\columnbreak
|
||||
Soit
|
||||
\begin{eqnarray*}
|
||||
2x + 4 = 0 \\
|
||||
2x + 4 \mathbf{+ (-4)} = \mathbf{-4}\\
|
||||
2x = -4 \\
|
||||
\mathbf{ \frac{1}{2} \times } 2x =\mathbf{\frac{1}{2} \times } -4 \\
|
||||
x = \frac{-4}{2} = -2
|
||||
\end{eqnarray*}
|
||||
|
||||
\end{multicols}
|
||||
Les solutions sont $x = \frac{-5}{3}$ ou $x = -2$.
|
||||
}}
|
||||
\end{center}
|
||||
|
||||
Résoudre l'équation $(6x + 2)(3x + 4) = 0$.
|
||||
|
||||
$(6x + 2)(3x + 4) = 0$ donc
|
||||
\begin{center}
|
||||
\parbox{3cm}{\dotfill} = 0\hspace{1cm} soit \hspace{1cm} \parbox{3cm}{\dotfill} = 0
|
||||
\end{center}
|
||||
|
||||
\begin{multicols}{2}
|
||||
Soit
|
||||
\begin{eqnarray*}
|
||||
\parbox{2cm}{\dotfill} = 0 \\[0.3cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.3cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.3cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.3cm]
|
||||
x = \parbox{1cm}{\dotfill}
|
||||
\end{eqnarray*}
|
||||
|
||||
\columnbreak
|
||||
Soit
|
||||
\begin{eqnarray*}
|
||||
\parbox{2cm}{\dotfill} = 0 \\[0.3cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.3cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.3cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.3cm]
|
||||
x = \parbox{1cm}{\dotfill}
|
||||
\end{eqnarray*}
|
||||
|
||||
\end{multicols}
|
||||
Les solutions sont $x = \parbox{1cm}{\dotfill}$ ou $x = $\parbox{1cm}{\dotfill}.
|
||||
\end{Exo}
|
||||
|
||||
|
||||
|
||||
|
||||
\end{document}
|
||||
|
||||
%%% Local Variables:
|
||||
%%% mode: latex
|
||||
%%% TeX-master: "master"
|
||||
%%% End:
|
||||
|
||||
Binary file not shown.
@@ -0,0 +1,132 @@
|
||||
\documentclass[a4paper,12pt,landscape, twocolumn]{/media/documents/Cours/Prof/Enseignements/Archive/2013-2014/tools/style/classExo}
|
||||
|
||||
\usepackage{multicol}
|
||||
|
||||
% Title Page
|
||||
\title{Identités remarquables et équations- Exercices}
|
||||
\author{}
|
||||
\date{}
|
||||
|
||||
\fancyhead[L]{Troisième}
|
||||
\fancyhead[C]{\Thetitle}
|
||||
\fancyhead[R]{\thepage}
|
||||
|
||||
|
||||
\begin{document}
|
||||
\thispagestyle{fancy}
|
||||
|
||||
|
||||
\begin{Exo}
|
||||
\exo{Équation produit}
|
||||
\begin{center}
|
||||
\framebox{\parbox{0.45\textwidth}{
|
||||
Résoudre l'équation $(3x + 5)(2x + 4) = 0$.
|
||||
|
||||
$(3x + 5)(2x + 4) = 0$ donc
|
||||
\begin{center}
|
||||
$3x + 5 = 0$ \hspace{1cm} soit \hspace{1cm} $2x + 4 = 0$
|
||||
\end{center}
|
||||
|
||||
\begin{multicols}{2}
|
||||
Soit
|
||||
\begin{eqnarray*}
|
||||
3x + 5 = 0 \\
|
||||
3x + 5 \mathbf{+ (-5)} = \mathbf{-5}\\
|
||||
3x = -5 \\
|
||||
\mathbf{\frac{1}{3} \times } 3x = \mathbf{ \frac{1}{3} \times } (-5) \\
|
||||
x = \frac{-5}{3}
|
||||
\end{eqnarray*}
|
||||
|
||||
\columnbreak
|
||||
Soit
|
||||
\begin{eqnarray*}
|
||||
2x + 4 = 0 \\
|
||||
2x + 4 \mathbf{+ (-4)} = \mathbf{-4}\\
|
||||
2x = -4 \\
|
||||
\mathbf{ \frac{1}{2} \times } 2x =\mathbf{\frac{1}{2} \times } -4 \\
|
||||
x = \frac{-4}{2} = -2
|
||||
\end{eqnarray*}
|
||||
|
||||
\end{multicols}
|
||||
Les solutions sont $x = \frac{-5}{3}$ ou $x = -2$.
|
||||
}}
|
||||
\end{center}
|
||||
|
||||
Résoudre l'équation $(2x + 2)(5x + 10) = 0$.
|
||||
|
||||
$(2x + 2)(5x + 10) = 0$ donc
|
||||
\begin{center}
|
||||
\parbox{4cm}{\dotfill} = 0\hspace{1cm} soit \hspace{1cm} \parbox{4cm}{\dotfill} = 0
|
||||
\end{center}
|
||||
|
||||
\begin{multicols}{2}
|
||||
Soit
|
||||
\begin{eqnarray*}
|
||||
\parbox{2cm}{\dotfill} = 0 \\[0.5cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.5cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.5cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.5cm]
|
||||
x = \parbox{1cm}{\dotfill}
|
||||
\end{eqnarray*}
|
||||
|
||||
\columnbreak
|
||||
Soit
|
||||
\begin{eqnarray*}
|
||||
\parbox{2cm}{\dotfill} = 0 \\[0.5cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.5cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.5cm]
|
||||
\parbox{2cm}{\dotfill} = \parbox{2cm}{\dotfill} \\[0.5cm]
|
||||
x = \parbox{1cm}{\dotfill}
|
||||
\end{eqnarray*}
|
||||
|
||||
\end{multicols}
|
||||
Les solutions sont $x = \parbox{1cm}{\dotfill}$ ou $x = $\parbox{1cm}{\dotfill}.
|
||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
Résoudre les équations suivantes
|
||||
\begin{multicols}{2}
|
||||
\begin{enumerate}
|
||||
\item $(2x + 3)(4x + 5) = 0$
|
||||
\item $(6x + 13)(44x + 5) = 0$
|
||||
\item $(12x - 4)(54x + 6) = 0$
|
||||
\item $(x - 3)(-4x - 2) = 0$
|
||||
\columnbreak
|
||||
|
||||
\item $(x - 1)(x - 2) = 0$
|
||||
\item $(2x + 3)^2 = 0$
|
||||
\item $(3x + 8)(3x - 8) = 0$
|
||||
\item $(7x - 1)^2 = 0$
|
||||
\end{enumerate}
|
||||
\end{multicols}
|
||||
\end{Exo}
|
||||
|
||||
\begin{Exo}
|
||||
Voici trois expressions
|
||||
\begin{eqnarray*}
|
||||
A = 9x^2 - 16 \hspace{1cm} B = 4x^2 + 24x + 9 \\
|
||||
C = (x + 1)(2x - 4) + (x + 1)(7x + 2)
|
||||
\end{eqnarray*}
|
||||
\begin{enumerate}
|
||||
\item Factoriser les trois expressions
|
||||
\item En utilisant la forme de l'énoncé, développer $C$.
|
||||
\item En utilisant la forme factorisée, résoudre les équations
|
||||
\begin{eqnarray*}
|
||||
A = 0 \hspace{1cm} B = 0 \hspace{1cm} C = 0
|
||||
\end{eqnarray*}
|
||||
|
||||
\end{enumerate}
|
||||
|
||||
\end{Exo}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\end{document}
|
||||
|
||||
%%% Local Variables:
|
||||
%%% mode: latex
|
||||
%%% TeX-master: "master"
|
||||
%%% End:
|
||||
|
||||
@@ -0,0 +1,37 @@
|
||||
Notes sur des fiches d'exercices autour des identites remarquables
|
||||
##################################################################
|
||||
|
||||
:date: 2014-07-01
|
||||
:modified: 2014-07-01
|
||||
:tags: Nombres Calculs, Identités Remarquables
|
||||
:category: 3e
|
||||
:authors: Benjamin Bertrand
|
||||
:summary: Pas de résumé, note créée automatiquement parce que je ne l'avais pas bien fait...
|
||||
|
||||
|
||||
|
||||
`Lien vers exo_id_rmq_eq_2.pdf <exo_id_rmq_eq_2.pdf>`_
|
||||
|
||||
`Lien vers exo_id_rmq_4.tex <exo_id_rmq_4.tex>`_
|
||||
|
||||
`Lien vers exo_id_rmq.pdf <exo_id_rmq.pdf>`_
|
||||
|
||||
`Lien vers exo_id_rmq.tex <exo_id_rmq.tex>`_
|
||||
|
||||
`Lien vers exo_id_rmq_4.pdf <exo_id_rmq_4.pdf>`_
|
||||
|
||||
`Lien vers exo_id_rmq_2.tex <exo_id_rmq_2.tex>`_
|
||||
|
||||
`Lien vers exo_id_rmq_3.pdf <exo_id_rmq_3.pdf>`_
|
||||
|
||||
`Lien vers exo_id_rmq_eq_2.tex <exo_id_rmq_eq_2.tex>`_
|
||||
|
||||
`Lien vers exo_id_rmq_eq_1.pdf <exo_id_rmq_eq_1.pdf>`_
|
||||
|
||||
`Lien vers exo_id_rmq_eq_1.tex <exo_id_rmq_eq_1.tex>`_
|
||||
|
||||
`Lien vers exo_id_rmq_2.pdf <exo_id_rmq_2.pdf>`_
|
||||
|
||||
`Lien vers exo_id_rmq_3.tex <exo_id_rmq_3.tex>`_
|
||||
|
||||
Il semblerait que la première étape qui consiste à relier les expressions justes aident les élèves.
|
||||
Reference in New Issue
Block a user