Commit 4bf3b8f7 by Louis

Mise à jour des tests

parent 45fcc747
 ... ... @@ -8,18 +8,18 @@ On considère le trinôme du second degré $f: x\mapsto \numprint{2}\,x^2 -\nump \item Soit$x\in\mathbb{R}. Alors : \begin{align*} \numprint{2} \,\left( x +\numprint{5} \right) \, \left( x -\numprint{7} \right) &= \numprint{2} \,\left(x\times x +\numprint{5}\times x -\numprint{7}\times x +\numprint{5}\times \left(\numprint{-7}\right) \right) \\ &= \numprint{2} \,\left(x\times{} x +\numprint{5}\times{} x -\numprint{7}\times{} x +\numprint{5}\times{} \left(\numprint{-7}\right) \right) \\ &= \numprint{2} \,\left(x^2 -\numprint{2}\,x -\numprint{35}\right) \\ &= \numprint{2}\,x^2 +\numprint{2}\times\left(\numprint{-2}\,x\right) +\numprint{2}\times\left(\numprint{-35}\right) \\ &= \numprint{2}\times{} x^2 +\numprint{2}\times{} \left(\numprint{-2}\,x\right) +\numprint{2}\times{} \left(\numprint{-35}\right) \\ &= \numprint{2}\,x^2 -\numprint{4}\,x -\numprint{70}\\ &= f\,(x) \end{align*} \item Soitx\in\mathbb{R}. Alors : \begin{align*} \numprint{2}\,\left( x -\numprint{1} \right)^2 -\numprint{72} &= \numprint{2}\,\left( x^2 -\numprint{2}\times \numprint{1} \times x + \numprint{1}^2\right) -\numprint{72}\\ &= \numprint{2}\,\left( x^2 -\numprint{2}\times{} \numprint{1} \times{} x + \numprint{1}^2\right) -\numprint{72}\\ &= \numprint{2}\,\left( x^2 -\numprint{2}\,x + \numprint{1}\right) -\numprint{72}\\ &= \numprint{2}\,x^2 +\numprint{2}\times \left(\numprint{-2}\,x\right) +\numprint{2}\times \numprint{1} -\numprint{72}\\ &= \numprint{2}\times{} x^2 +\numprint{2}\times{} \left(\numprint{-2}\,x\right) +\numprint{2}\times{} \numprint{1} -\numprint{72}\\ &= \numprint{2}\,x^2 -\numprint{4}\,x +\numprint{2} -\numprint{72}\\ &= \numprint{2}\,x^2 -\numprint{4}\,x -\numprint{70}\\ &= f\,(x) ... ... @@ -43,7 +43,7 @@ f\,(x) &= \numprint{-70} \\ Nous pouvons maintenant factoriser le membre de gauche parx, ce qui nous donnera une équation produit nul. \begin{align*} \numprint{2}\,x^2 -\numprint{4}\,x &= 0 \\ \numprint{2}\,x\times x -\numprint{4}\times x &= 0 \\ \numprint{2}\,x\times{} x -\numprint{4}\times{} x &= 0 \\ x\,\left( \numprint{2}\,x -\numprint{4} \right) &= 0 \\ \end{align*} \begin{align*} ... ... @@ -74,8 +74,8 @@ Il y a donc une unique solutionx=\numprint{1}$. positif , donc la fonction est décroissante puis croissante . De plus, l'absisse du sommet est$-\frac{\numprint{-4}}{2\times\numprint{2}}$, soit$\numprint{1}$, et$f\,( \numprint{1} )=\numprint{2}\times\numprint{1}^2 -\numprint{4}\times \numprint{1} -\numprint{70}=\numprint{-72}$. De plus, l'absisse du sommet est$-\frac{\numprint{-4}}{2\times{}\numprint{2}}$, soit$\numprint{1}$, et$f\,( \numprint{1} )=\numprint{2}\times{}\numprint{1}^2 -\numprint{4}\times{} \numprint{1} -\numprint{70}=\numprint{-72}$. Le tableau de variations est donc : \begin{center} \begin{tikzpicture} ... ...  ... ... @@ -8,18 +8,18 @@ On considère le trinôme du second degré$f: x\mapsto \numprint{-2}\,x^2 -\num \item Soit $x\in\mathbb{R}$. Alors : \begin{align*} \numprint{-2} \,\left( x -\numprint{5} \right) \, \left( x +\numprint{8} \right) &= \numprint{-2} \,\left(x\times x -\numprint{5}\times x +\numprint{8}\times x -\numprint{5}\times \numprint{8} \right) \\ &= \numprint{-2} \,\left(x\times{} x -\numprint{5}\times{} x +\numprint{8}\times{} x -\numprint{5}\times{} \numprint{8} \right) \\ &= \numprint{-2} \,\left(x^2 +\numprint{3}\,x -\numprint{40}\right) \\ &= \numprint{-2}\,x^2 -\numprint{2}\times\numprint{3}\,x -\numprint{2}\times\left(\numprint{-40}\right) \\ &= \numprint{-2}\times{} x^2 -\numprint{2}\times{} \numprint{3}\,x -\numprint{2}\times{} \left(\numprint{-40}\right) \\ &= \numprint{-2}\,x^2 -\numprint{6}\,x +\numprint{80}\\ &= f\,(x) \end{align*} \item Soit $x\in\mathbb{R}$. Alors : \begin{align*} \numprint{-2}\,\left( x +\numprint{1.5} \right)^2 +\numprint{84.5} &= \numprint{-2}\,\left( x^2 +\numprint{2}\times \numprint{1.5} \times x + \numprint{1.5}^2\right) +\numprint{84.5}\\ &= \numprint{-2}\,\left( x^2 +\numprint{2}\times{} \numprint{1.5} \times{} x + \numprint{1.5}^2\right) +\numprint{84.5}\\ &= \numprint{-2}\,\left( x^2 +\numprint{3}\,x + \numprint{2.25}\right) +\numprint{84.5}\\ &= \numprint{-2}\,x^2 -\numprint{2}\times \numprint{3}\,x -\numprint{2}\times \numprint{2.25} +\numprint{84.5}\\ &= \numprint{-2}\times{} x^2 -\numprint{2}\times{} \numprint{3}\,x -\numprint{2}\times{} \numprint{2.25} +\numprint{84.5}\\ &= \numprint{-2}\,x^2 -\numprint{6}\,x -\numprint{4.5} +\numprint{84.5}\\ &= \numprint{-2}\,x^2 -\numprint{6}\,x +\numprint{80}\\ &= f\,(x) ... ... @@ -43,7 +43,7 @@ f\,(x) &= \numprint{80} \\ Nous pouvons maintenant factoriser le membre de gauche par $x$, ce qui nous donnera une équation produit nul. \begin{align*} \numprint{-2}\,x^2 -\numprint{6}\,x &= 0 \\ \numprint{-2}\,x\times x -\numprint{6}\times x &= 0 \\ \numprint{-2}\,x\times{} x -\numprint{6}\times{} x &= 0 \\ x\,\left( \numprint{-2}\,x -\numprint{6} \right) &= 0 \\ \end{align*} \begin{align*} ... ... @@ -74,8 +74,8 @@ Il y a donc une unique solution $x=\numprint{-1.5}$. négatif , donc la fonction est croissante puis décroissante . De plus, l'absisse du sommet est $-\frac{\numprint{-6}}{2\times\left(\numprint{-2}\right)}$, soit $\numprint{-1.5}$, et $f\,( \numprint{-1.5} )=\numprint{-2}\times\left(\numprint{-1.5}\right)^2 -\numprint{6}\times \left(\numprint{-1.5}\right) +\numprint{80}=\numprint{84.5}$. De plus, l'absisse du sommet est $-\frac{\numprint{-6}}{2\times{}\left(\numprint{-2}\right)}$, soit $\numprint{-1.5}$, et $f\,( \numprint{-1.5} )=\numprint{-2}\times{}\left(\numprint{-1.5}\right)^2 -\numprint{6}\times{} \left(\numprint{-1.5}\right) +\numprint{80}=\numprint{84.5}$. Le tableau de variations est donc : \begin{center} \begin{tikzpicture} ... ...
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