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+\documentclass[a4paper,12pt]{classPres}
+\usepackage{tkz-fct}
+
+\author{}
+\title{}
+\date{}
+
+\begin{document}
+\begin{frame}{Questions flashs}
+    Développer  et réduire l'expression
+    \[ A = (x-3)\times4 - (x-3)^2\]
+    Mettre sous forme canonique
+    \[ B = 9x^2 - 54x + 65 \]
+    \textbf{Bonus} factoriser ces 2 expressions.
+\end{frame}
+
+\begin{frame}{Correction}
+    \begin{eqnarray*}
+        A &=& 4x - 12 - (x^2 - 6x +9) \\
+          &=& 4x - 12 - x^2 + 6x - 9 \\
+          &=& -x^2 + 10x - 21
+    \end{eqnarray*}
+    \hline
+    \begin{eqnarray*}
+       B = 9x^2 - 54x + 65
+    \end{eqnarray*}
+    \[ \alpha = \frac{-b}{2a} = 3 \]
+    \[ \beta = -\frac{b^2-4ac}{4a} = -\frac{36+38}{4} = -16 \]
+    \begin{eqnarray*}
+        B &=& 9(x - 3)^2 - 16 \\
+    \end{eqnarray*}
+\end{frame}
+\begin{frame}{Bonus}
+    \begin{eqnarray*}
+        A &=& (x-3)\times4 - (x-3)^2\\
+          &=& (x-3)(4 - x + 3)\\
+          &=& (x-3)(-x + 7)
+    \end{eqnarray*}
+    \hline
+    \begin{eqnarray*}
+        B &=& 9(x - 3)^2 - 16 \\
+          &=& \left[ 3(x-3) \right]^2- 4^2 \\
+          &=& ( 3x - 9 - 4 ) (3x - 9 + 4) \\
+          &=& (3x - 13)(3x - 5)
+    \end{eqnarray*}
+
+\end{frame}
+
+\end{document}