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\begin{center}\section*{\textcolor{red}{Correction des exercices sur les identités remarquables}}\end{center}




\subsection{}
Développer en utilisant les identités remarquables :

\begin{enumerate}[$A=$]


\bigskip

\item  $(x-5)^2x^2-2\times x\times 5+5^2=\boxed{\textcolor{red}{x^2-10x+25}}$


\bigskip

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


\bigskip

\item  $\left(\dfrac{1}{2}x+1\right)^2=\left(\dfrac{1}{2}x\right)^2+2\times \dfrac{1}{2}x\times 1+1^1=\boxed{\textcolor{red}{\dfrac{1}{4}x^2+x+1}}$

\bigskip

\item  $(2x - 7) (2x + 7)=(2x)^2-7^2=\boxed{\textcolor{red}{4x^2-49}}$

\bigskip

\item  $\left(\dfrac{1}{3}x-4\right)\left(\dfrac{1}{3}x+4\right)=\left(\dfrac{1}{3}x\right)^2-4^2=\boxed{\textcolor{red}{\dfrac{1}{9}x^2-16}}$

\bigskip

\item  $\left(2x-\sqrt{3}\right)\left(2x+\sqrt{3}\right)=(2x^2)-\sqrt{3}^2=\boxed{\textcolor{red}{4x^2-3}}$

\bigskip

\item  $\left(x+\dfrac{1}{x}\right)^2=x^2+2\times x\times \dfrac{1}{x}+\left(\dfrac{1}{x}\right)^2=\boxed{\textcolor{red}{x^2+2+\dfrac{1}{x^2}}}$

\bigskip

\item  $\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{3}^2-\sqrt{2}^2=3-2=\boxed{\textcolor{red}{1}}$

\bigskip

\item  $(7-2x)(2x-7)=-(2x-7)(2x+7)=-\left[(2x)^2-7^2\right]=-\left(4x^2-49\right)=\boxed{\textcolor{red}{49-4x^2}}$
\end{enumerate}

\subsection{}
Développer et simplifier les expressions suivantes :
\begin{enumerate}[$A=$]

\bigskip

\item   $\left(\sqrt{7}- 3\right)\left(\sqrt{7}+3\right)=\sqrt{7}^2-3^2=7-9=\boxed{\textcolor{red}{-2}}$

\bigskip

\item  $\left(\sqrt{3}+\sqrt{5}\right)^2+\left(\sqrt{15}-1\right)^2\\
=\left[\sqrt{3}^2+2\times \sqrt{3}\times \sqrt{5}+\sqrt{5}^2\right]+\left[\sqrt{15}^2-2\times \sqrt{15}\times 1+1^2\right]\\
=\left[3+2\sqrt{15}+5\right]+\left[15-2\sqrt{15}+1\right]\\
=8+2\sqrt{15}+16-2\sqrt{15}=\boxed{\textcolor{red}{24}}$

\bigskip

\item  $\left(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\right)^2\\
=\sqrt{4-\sqrt{7}}^2+2\times \sqrt{4-\sqrt{7}}\times \sqrt{4+\sqrt{7}}+\sqrt{4+\sqrt{7}}^2\\
=4-\sqrt{7}+2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\\
=4-\sqrt{7}+4+\sqrt{7}+2\sqrt{4^2-\sqrt{7}^2}\\
=8+2\sqrt{16-7}=8+2\sqrt{9}=8+2\times 3=\boxed{\textcolor{red}{14}}$
\end{enumerate}

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