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\begin{document}

\titreCours{5}{Produits remarquables}{}{}

\begin{frame}{1. Formules du carré}

\begin{rframed}
\begin{eqnarray*}
(A\bl{+}B)^{\ve{2}} & = \uncover<2->{&  A^2 \bl{+} \ve{2}AB + B^2 } \\
\uncover<3->{(A\ro{-}B)^{\ve{2}} & = & \uncover<4->{A^2 \ro{-} \ve{2}AB + B^2}}\\
\uncover<5->{(A\ro{-}B)\cdot (A\bl{+}B) & = & \uncover<6->{A^2 \ro{-} B^2}}\\
\end{eqnarray*}
\end{rframed}

\vspace{0.5cm}

\hunc{2}{
\uncover<7->{
Démonstration de $(A+B)^2 = A^2 + 2AB + B^2$
\begin{eqnarray*}
(A + B)^2 
\uncover<8->{&=&} \only<8>{(A + B)(A + B)}
\only<9-10>{(\bl{A} + B)(\bl{A} + B)}
\only<11-12>{(\bl{A} + B)(A + \bl{B})}
\only<13-14>{(A + \bl{B})(\bl{A} + B)}
\only<15->{(A + \bl{B})(A + \bl{B})}
\\
\uncover<10->{&=& A^2}\uncover<12->{+ AB}\uncover<14->{+ BA} \uncover<15->{+ B^2}\\
\uncover<16->{&=& A^2 + AB + \ro{AB} + B^2}\\
\uncover<17->{&=& A^2 + 2AB + B^2}
\end{eqnarray*}

}}

\end{frame}

\begin{frame}{$\textcolor{black}{(\bl{A}+\ve{B})^2 = \bl{A}^2 + 2\bl{A}\ve{B} + \ve{B}^2}$}

\underline{Exemple 1.1} Développer: $(3x + 4y)^2$\\
\vfill
\hunc{2}{
\uncover<2->{On a donc $\bl{A = \uncover<3->{3x}}$ et $\ve{B=\uncover<3->{4y}}$:
\begin{enumerate}
\item<4-> $\bl{A}^2 
\uncover<5->{= (\bl{3x})^2}
\uncover<6->{= 3^2 \cdot x^2}
\uncover<7->{= 9 \cdot x^2}
\uncover<9->{= \boxed{9x^2}}
$\\[0.2cm]
\item<10-> $2\bl{A}\ve{B} 
\uncover<11->{= 2 \cdot (\bl{3x}) \cdot (\ve{4y})}
\uncover<12->{= 2 \cdot 3 \cdot x \cdot 4 \cdot y}
\uncover<13->{= 2 \cdot 3 \cdot 4 \cdot x \cdot y}$\\
$\quad\quad\uncover<14->{= 24 \cdot x \cdot y}
\uncover<15->{= \boxed{24xy}}
$\\[0.2cm]
\item<16-> $\ve{B}^2 
\uncover<17->{= (\ve{4y})^2}
\uncover<18->{= 4^2 \cdot y^2}
\uncover<19->{= 16 \cdot y^2}
\uncover<20->{= \boxed{16y^2}}
$\\[1cm]
\end{enumerate}


\uncover<21->{On obtient donc:
$$(\bl{3x}+\ve{4y})^2 = \uncover<22->{9x^2} + \uncover<23->{24xy} + \uncover<24->{16y^2} $$
}
}
}

\end{frame}


\begin{frame}{$\textcolor{black}{(\bl{A}\ro{-}\ve{B})^2 = \bl{A}^2 \ro{-} 2\bl{A}\ve{B} + \ve{B}^2}$}

\underline{Exemple 1.2} Développer: $(x^2y - 5x)^2$\\
\vfill

\hunc{2}{
\uncover<2->{On a donc $\bl{A = \uncover<3->{x^2y}}$ et $\ve{B=\uncover<3->{5x}}$:
\begin{enumerate}
\item<4-> $\bl{A}^2 
\uncover<5->{= (\bl{x^2y})^2}
\uncover<6->{= (x^2)^2 \cdot y^2}
\uncover<7->{= x^4 \cdot y^2}
\uncover<9->{= \boxed{x^4y^2}}
$\\[0.2cm]
\item<10-> $2\bl{A}\ve{B} 
\uncover<11->{= 2 \cdot (\bl{x^2y}) \cdot (\ve{5x})}
\uncover<12->{= 2 \cdot x^2 \cdot y \cdot 5 \cdot x}
\uncover<13->{= 2 \cdot 5 \cdot x^2 \cdot x \cdot y}$\\
$\qquad\qquad\uncover<14->{= 10 \cdot x^3 \cdot y}
\uncover<15->{= \boxed{10x^3y}}
$\\[0.2cm]
\item<16-> $\ve{B}^2 
\uncover<17->{= (\ve{5x})^2}
\uncover<18->{= 5^2 \cdot x^2}
\uncover<19->{= 25 \cdot x^2}
\uncover<20->{= \boxed{25x^2}}
$\\[0.5cm]
\end{enumerate}


\uncover<21->{On obtient donc:
$$(\bl{x^2y}\ro{-}\ve{5x})^2 = \uncover<22->{x^4y^2} \ro{-} \uncover<23->{10x^3y} + \uncover<24->{25x^2} $$
}}}

\end{frame}

\begin{frame}{\textcolor{black}{$(\bl{A}+\ve{B})(\bl{A}-\ve{B})=\bl{A}^2-\ve{B}^2$}}

\underline{Exemple 1.3} Développez: $\left(\bl{\dfrac{4x}{5}}+\ve{\dfrac{3y}{4}}\right)\left(\bl{\dfrac{4x}{5}}-\ve{\dfrac{3y}{4}}\right)$ \\
\vfill
\small

\hunc{2}{
\uncover<2->{On a donc $\bl{A = \uncover<3->{\dfrac{4x}{5}}}$ et $\ve{B=\uncover<3->{\dfrac{3y}{4}}}$:
\begin{enumerate}
\item<4-> $\bl{A}^2 
\uncover<5->{= \left(\bl{\dfrac{4\cdot x}{5}}\right)^2}
\uncover<6->{= \dfrac{4^2 \cdot x^2}{5^2}}
\uncover<7->{= \dfrac{16 \cdot x^2}{25}}
\uncover<8->{= \dfrac{16}{25} \cdot x^2}
\uncover<9->{= \boxed{\dfrac{16}{25}x^2}}
$\\[0.2cm]
\item<10-> $\ve{B}^2 
\uncover<11->{= \left(\ve{\dfrac{3\cdot y}{4}}\right)^2}
\uncover<12->{= \dfrac{3^2\cdot y^2}{4^2}}
\uncover<13->{= \dfrac{9\cdot y^2}{16}}
\uncover<14->{= \boxed{\dfrac{9}{16}y^2}}
$\\[0.5cm]
\end{enumerate}


\uncover<15->{On obtient donc:
$$\left(\bl{\dfrac{4x}{5}}+\ve{\dfrac{3y}{4}}\right)\left(\bl{\dfrac{4x}{5}}-\ve{\dfrac{3y}{4}}\right) 
	= \uncover<16->{\dfrac{16}{25}x^2} \ro{-} \uncover<17->{\dfrac{9}{16}y^2}$$
}}}

\end{frame}

\begin{frame}{2. Formules du cube}

\small

\begin{cframed}
\begin{eqnarray*}
(A\bl{+}B)^{\ve{3}} & = & \uncover<2->{ A^{\ve{3}} \bl{+} \ve{3} A^2 B \bl{+} \ve{3} AB^2 \bl{+} B^{\ve{3}}}\\
\uncover<3->{(A\ro{-}B)^{\ve{3}} & = & \uncover<4->{A^{\ve{3}} \ro{-} \ve{3} A^2 B \bl{+} \ve{3} AB^2 \ro{-} B^{\ve{3}}}}\\
\uncover<5->{(A \bl{+} B)(A^2 \ro{-} AB + B^2) & = & \uncover<6->{A^{\ve{3}} \bl{+} B^{\ve{3}}}}\\
\uncover<7->{(A \ro{-} B)(A^2 \bl{+} AB + B^2) & = & \uncover<8->{A^{\ve{3}} \ro{-} B^{\ve{3}}}}\\[0.5cm]
\end{eqnarray*}
\end{cframed}

\end{frame}


\begin{frame}{$\textcolor{black}{(\bl{A}\ro{-}\ve{B})^{3} = \bl{A}^{3} \ro{-} 3 \bl{A}^2 \ve{B} \bl{+} 3 \bl{A}\ve{B}^2 \ro{-} \ve{B}^{3}}$}

\underline{Exemple 2.1} Développez $(4x - 3y)^3$\\
\vfill
\small


\hunc{2}{
\uncover<2->{
On a $\bl{A= \uncover<3->{4x}}$ et $\ve{B= \uncover<4->{3y}}$.\\[0.2cm]

\uncover<5->{
On calcule les différents termes:\\

\begin{enumerate}
\item<5-> $\bl{A}^3 
\uncover<6->{= (\bl{4x})^3} 
\uncover<7->{= 4^3 \cdot x^3}
\uncover<8->{= \boxed{64x^3}}$\\[0.2cm]
\item<9-> $3\bl{A}^2 \ve{B} 
\uncover<10->{= 3\cdot (\bl{4x})^2\cdot \ve{3y}} 
\uncover<11->{= 3 \cdot 4^2 \cdot x^2 \cdot 3 \cdot y}
\uncover<12->{= 3 \cdot 16 \cdot x^2 \cdot 3 \cdot y}$\\
$\quad\quad\uncover<13->{= 144 \cdot x^2 \cdot y}
\uncover<14->{= \boxed{144x^2 y}}$\\[0.2cm]
\item<15-> $3\bl{A}\ve{B}^2 
\uncover<16->{= 3 \cdot \bl{4x} \cdot (\ve{3y})^2} 
\uncover<17->{= 3 \cdot 4 \cdot x \cdot 3^2 \cdot y^2}
\uncover<18->{= 3 \cdot 4 \cdot x \cdot 9 \cdot y^2}$\\
$\quad\quad\uncover<19->{= 108 \cdot x \cdot y^2}
\uncover<20->{= \boxed{108xy^2}}$\\[0.2cm]
\item<21-> $\ve{B}^3 
\uncover<22->{= (\ve{3y})^3} 
\uncover<23->{= 3^3 \cdot y^3}
\uncover<24->{= \boxed{27y^3}}$\\[0.5cm]
\end{enumerate}}
\uncover<25->{
On obtient donc:
$$(4x \ro{-} 3y)^3 = \uncover<26->{64x^3 \ro{-} 144x^2y + 108xy^2 \ro{-} 27y^3}$$}

}}

\end{frame} 


\begin{frame}{$\textcolor{black}{(\bl{A} \ro{-} \ve{B})(\bl{A}^2 + \bl{A}\ve{B} + \ve{B}^2) = \bl{A}^3 \ro{-} \ve{B}^{3}}$}

\underline{Exemple 2.2} Développez $(3x - 5y)(9x^2 + 15xy + 25y^2)$\\
\vfill
\small
\hunc{2}{
\uncover<2->{
On a $\bl{A= \uncover<3->{3x}}$ et $\ve{B= \uncover<4->{5y}}$.\\[0.2cm]

\uncover<5->{
On vérifie le deuxième facteur: $A^2= \uncover<6->{9x^2}$, $AB=\uncover<7->{15xy}$ et $B^2=\uncover<8->{25y^2}$. \\[0.2cm]
}

\uncover<9->{
On calcule les différents termes:\\

\begin{enumerate}
\item<10-> $\bl{A}^3 
\uncover<11->{= (\bl{3x})^3} 
\uncover<12->{= 3^3 \cdot x^3}
\uncover<13->{= \boxed{27x^3}}$\\[0.2cm]
\item<14-> $\ve{B}^3 
\uncover<15->{= (\ve{5y})^3} 
\uncover<16->{= 5^3 \cdot y^3}
\uncover<17->{= \boxed{125y^3}}$\\[0.5cm]
\end{enumerate}}
\uncover<18->{
On obtient donc:
$$(3x - 5y)(9x^2 + 15xy + 25y^2) \uncover<19->{= 27x^3 - 125y^3}$$}

}}
\end{frame} 


\end{document}










