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\begin{document}%%%%%%%%%%b{gedeonInfo}%%%%%%%%%
%%--Document N° 1508212587621--
%%--Corrigé du document N° 1508216786258--
%%%%%%%%%%e{gedeonInfo}%%%%%%%%%



\begin{claim}
$\quad$\vspace*{0.35cm} 

\begin{enumerate}


\item
\begin{enumerate}


\item $X=\dfrac{\left(  -a^{2}b^{3}\right) ^{3}}{\left(  -a^{3}b\right) ^{2}%
}=\dfrac{-a^{6}b^{9}}{a^{6}b^{2}}=-a^{0}b^{7}=-b^{7}$. 

\item $Y=\dfrac{\left(  -ab\right) ^{-1}}{\left(  -a^{-3}b^{3}\right) ^{2}%
}\times a^{-3}b^{3}=-a^{-1}b^{-1}\times a^{6}b^{-6}\times a^{-3}b^{3}%
=-a^{2}b^{-4}$. 
\end{enumerate}

\item $\left( \sqrt{3}-2\sqrt{2}\right) ^{2}=3-4\sqrt{6}+8=11-4\sqrt{6}%
$\newline donc  $\sqrt{11-4\sqrt{6}}=\sqrt{\left( \sqrt{3}-2\sqrt{2}\right)
^{2}}=\left| \sqrt{3}-2\sqrt{2}\right| =2\sqrt{2}-\sqrt{3}\quad$ car $\sqrt
{3}-2\sqrt{2}<0$. 

\item
\begin{align*}
-2x^{2}+3x-1= -2\left( x^{2}-\frac{3}{2}x+\frac{1}{2}\right)  & =-2\left(
\left( x-\frac{3}{4}\right) ^{2}-\frac{9}{16}+\frac{8}{16}\right) \\
& =-2\left( \left( x-\frac{3}{4}\right) ^{2}-\frac{1}{4}\right) =-2\left(
\left( x-\frac{3}{4}-\frac{1}{4}\right) \left( x-\frac{3}{4}+\frac{1}%
{4}\right) \right) \\
& =-2\left( x-1\right) \left( x-\frac{1}{2}\right) =-(x-1)(2x-1)
\end{align*}
$f(x)=\sqrt{-2x^{2}+3x-1}$ existe ssi $-2x^{2}+3x-1\geq0$ i.e ssi
$-(x-1)(2x-1)\geq0$.\newline D'aprés le tableau de signes suivant\newline
\begin{tabular}
[c]{c|ccc|ccc|ccc}%
$x$ & $-\infty$ &  & \multicolumn{2}{c}{$\frac{1}{2}$} &  &
\multicolumn{2}{c}{$1$} &  & $+\infty$\\\hline
&  &  &  &  &  &  &  &  & \\
$x-1$ &  & $-$ &  &  & $-$ & \multicolumn{2}{c}{$0$} & $+$ & \\
&  &  &  &  &  &  &  &  & \\\hline
&  &  &  &  &  &  &  &  & \\
$2x-1$ &  & $-$ & \multicolumn{2}{c}{$0$} & $+$ &  &  & $+$ & \\
&  &  &  &  &  &  &  &  & \\\hline
&  &  &  &  &  &  &  &  & \\
$-(x-1)(2x-1)$ &  & $-$ & \multicolumn{2}{c}{$0$} & $+$ &
\multicolumn{2}{c}{$0$} & $-$ & \\
&  &  &  &  &  &  &  &  & \\\hline
\end{tabular}


on peut conclure que $\mathcal{D}_{f}=\left[ \frac{1}{2}\,;1\right] $.

\item
\begin{enumerate}


\item
\begin{align*}
A & =(x-2)(x+2)^{2}-(x^{2}-4)(3x+6)\\
& =(x-2)(x+2)^{2}-3(x-2)(x+2)(x+2)\\
& =(x-2)(x+2)^{2}(x+2-3)\\
& =(x+2)^{2}(x-2)(x-1)
\end{align*}


\item
\begin{align*}
B  & =\left(  x-x\sqrt{5}\right)  ^{2}-\left(  1+x\sqrt{5}\right)  ^{2}\\
& =x^{2}-2x^{2}\sqrt{5}+5x^{2}-\left(  1+2\sqrt{5}x+5x^{2}\right)  \\
& =\left(  1+5-2\sqrt{5}-5\right)  x^{2}-2\sqrt{5}x-1\\
& =\left(  1-2\sqrt{5}\right)  x^{2}-2\sqrt{5}x-1.
\end{align*}

\end{enumerate}
\end{enumerate}
\end{claim}


\end{document}