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\begin{document}


\begin{center}\section*{\textcolor{red}{ Correction de l'accompagnement spécialisé : développements-identités remarquables-séance séance 13 du 19 décembre}}\end{center}




\subsection{\textcolor{blue}{Distributivité}}
Développer :

\begin{enumerate}
\item 
\begin{enumerate}
\item $5x \times 6y =\boxed{\textcolor{red}{30xy}}$

\item $(6+x) \times 3x =3x(x+6)=\boxed{\textcolor{red}{3x^2+18x}}$

\item $5x(x+8) =\boxed{\textcolor{red}{5x^2+40x}}$

\item $(5 + y)( 6+ y ) =30+5y+6y+y^2=\boxed{\textcolor{red}{y^2+11y+30}}$
\end{enumerate}

\item 
\begin{enumerate}
\item $(7+ x )^2 =7^2+2\times 7\times x+x^249+14y+x^2=\boxed{\textcolor{red}{x^2+14x+49}}$

\item $(7x)^2=7^2\times x^2=\boxed{\textcolor{red}{49x^2}}$

\item $(7-x )^2 =\boxed{\textcolor{red}{x^2-14x+49}}$

\item $(x -7)^2 =\boxed{\textcolor{red}{x^-14x+49}}$ car $(x-7)^2=\left[-(7-x)^2\right]=(-1)^2\times (7-x)^2=(7-x)^2$
\end{enumerate}

\item 
\begin{enumerate}
\item $4(t-6)+3(5-t) =4t-24+15-3t=\boxed{\textcolor{red}{t-9}}$

\item $10\left(\dfrac{1}{5}x+3\right)-15x=\dfrac{10}{5}x+30-15x=\dfrac{2\times \cancel{5}}{\cancel{5}}x+30-15x=2x+30-15x=\boxed{\textcolor{red}{-13x+30}}$

\item $\dfrac{5}{3}\left(-6+\dfrac{3}{2}\right)=-\dfrac{5}{\cancel{3}}\times 2\times \cancel{3}+\dfrac{5}{\cancel{3}}\times \dfrac{\cancel{3}}{2}=-10+\dfrac{5}{2}=-\dfrac{20}{2}+\dfrac{5}{2}=\boxed{\textcolor{red}{-\dfrac{15}{2}}}$

\item $5\left(x+\dfrac{1}{3}\right)-4(x+3)=5x+\dfrac{5}{3}-4x-12=x+\dfrac{5}{3}-\dfrac{36}{3}=\boxed{\textcolor{red}{x-\dfrac{31}{3}}}$
\end{enumerate}

\end{enumerate}



\subsection{\textcolor{blue}{Identités remarquables}}
Développer :
\begin{enumerate}[a)]
\item $(t-5)^2=\boxed{\textcolor{red}{t^2-10t+25}}$

\item $(x+7)^2 =\boxed{\textcolor{red}{x^2+14x+49}}$

\item $(2 y-6)^2 =(2y)^2-2\times 2y\times 6+6^2=\boxed{\textcolor{red}{4y^2-24y+36}}$

\item $(y + 3)( y - 3 )=\boxed{\textcolor{red}{y^2-9}}$

\item $(3y+7)^2 =(3y)^2+2\times 3y\times 7+7^2=\boxed{\textcolor{red}{9y^2+42y+49}}$

\item $(4x-3)^2 =(4x)^2-2\times 4x\times 3+3^2=\boxed{\textcolor{red}{16x^2-24x+9}}$

\item $(2t+5)(2t-5) =(2t)^2-5^2=\boxed{\textcolor{red}{4t^2-25}}$

\item $(x^2+ 5 )(x^2 - 5) =\left(x^2\right)^2-5^2=\boxed{\textcolor{red}{x^4-25}}$

\end{enumerate}



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