 \input{../preambules/preCorrection.tex}
\usepackage{enumitem}

\begin{document}

\chead{Correction Calculs A5}
\lfoot{}
\cfoot{Calcul}

\begin{center}
\fbox{ \Large Correction Calculs du 03/08 au 09/08}
\end{center}
\vspace{30pt}

\setcounter{theorem}{28}

\begin{exo}[Calcul de puissances]
Exprimer les calculs suivants sous la forme la plus simple possible. ($n$ est un entier).
\begin{qmulti}{2}
\item $1^n =  \boxed{1}$
\item $(-1)^{2n+1} = (-1)^{2n} \times (-1) = \boxed{-1}$
\item $2^n - 2^{n-1} = 2^{n-1}(2 - 1) = \boxed{2^{n-1}}$
\item $3^n + 3^n + 3^n = 3^{n} (1 + 1 + 1) = \boxed{3^{n+1}}$
\end{qmulti}
\end{exo}

\begin{exo}[Simplification]

�crire les expressions sous la forme $x^a$. 
\def\columnseprulecolor{\color{white}}
\begin{multicols}{4}
\begin{enumerate}
\item $(2^5)^4 = 2^{20}$. 
\item $\dfrac{1}{4^3} = 4^{-3}$. 
\item $3^4 \times 3^{-2} = 3^2$. 
\item $\dfrac{4^3}{5^2}$ impossible.
\item $3^5 \times 7^5 = 21^5$.
\item $\dfrac{4^3}{4^{-2}} = 4^5$.
\item $3^6 \times 5^4$ impossible. 
\item $\dfrac{9^{-2}}{3^{-2}} = \dfrac{3^{-4}}{3^{-2}} = 3^{-2}$.
\end{enumerate}
\end{multicols}
\end{exo}

\begin{exo}[Simplification]
Simplifier (si possible) les expressions au maximum. 

\reponseq{
9 (-3)^{2n} & = 3^2 \times (-1)^{2n} \times 3^{2n} \\
			& = 1 \times 3^{2n+2} \\
			& = \boxed{3^{2(n+1)}}
}{
2^n 4^{n-3} & = 2^n \times (2^2)^{n-3} \\
			& = 2^n \times 2^{2n - 6} \\
			& = \boxed{2^{3n - 6}}
}{ 
(2^3)^2 & = 8^2 \\
 & = \boxed{64}
}{
4^{n-1}3^{2n-2} & = 4^{n-1} \times  (3^{2})^{n-1} \\
                & = 4^{n-1} \times  9^{n-1} \\
                & = \boxed{36^{n-1}}
}
\end{exo}

\begin{exo}[Puissances n�gative]
�crire les expressions sous la forme $x^a$. 

\reponset{ 
\dfrac{1}{x^{4-n}} & = x^{-(4-n)} \\
& = \boxed{x^{n-4}}
}{ 
\displaystyle \dfrac{x^2}{x^n} & = \boxed{x^{2-n}} 
}{
\displaystyle \frac{x^n \times y^n}{(xy)^4} & = \dfrac{(xy)^n}{(xy)^4} \\
& = \boxed{(xy)^{n-4}}
}

�crire les expressions sous la forme d'une fraction. 
\def\columnseprulecolor{\color{white}}
\begin{multicols}{3}
\begin{enumerate}[start=4]
\item \fbox{$\displaystyle x^{-5} = \dfrac{1}{x^5}$ }
\item \fbox{ $\displaystyle x^{n-4} = \dfrac{x^n}{x^4}$} 
\item \fbox{$\displaystyle x^{2-n} = \dfrac{x^2}{x^n}$ }
\end{enumerate}
\end{multicols}
\end{exo}


\begin{exo}[Puissances fractionnaires]
�crire les expressions sous la forme $x^a$. 
\def\columnseprulecolor{\color{white}}
\begin{multicols}{3}
\begin{enumerate}
\item \fbox{$\displaystyle x \sqrt{x} = x^{\frac{3}{2}}$.} 
\item \fbox{$\displaystyle \frac{1}{\sqrt{x}} = x^{- \frac{1}{2}}$.} 
\item \fbox{$\displaystyle \frac{x^4}{\sqrt{x}} = x^{\frac{7}{2}}$. }
\end{enumerate}
\end{multicols}
�crire les expressions en utilisant le symbole $\sqrt{\quad}$. 
\def\columnseprulecolor{\color{white}}
\begin{multicols}{3}
\begin{enumerate}[start=4]
\item \fbox{$\displaystyle x^{\frac{3}{2}} = x \sqrt{x}$ }
\item \fbox{$\displaystyle x^{-\frac{1}{2}} = \dfrac{1}{\sqrt{x}}$} 
\item \fbox{$\displaystyle x^{\frac{5}{2}} = x^2 \sqrt{x}$ }
\end{enumerate}
\end{multicols}
\end{exo}

\newpage
\begin{exo}[Calculs]
�cris les expressions suivantes sous la forme d'un produit de puissances de nombres entiers, ayant le moins de facteurs possible. Tu d�tailleras les �tapes de calculs.

\reponset{ 
A & = \displaystyle \frac{3^4\times 2^5 \times 5^6}{3^7 \times 2^9 \times 5^3} \\
  & = \boxed{3^{-3} \times 2^{-4} \times 3^{3}}
}{
B & = \displaystyle \frac{7^{12}\times (9^4)^3 \times 5^{-5}}{9^{10} \times (5^{-7})^6 \times 7^{-17}} \\
  & = \dfrac{7^{29} \times 9^{12} \times 5^{-5}}{9^{10} \times 5^{-42} } \\
  & = \boxed{7^{29} \times 9^2 \times 5^{37} }
}{
C & = \displaystyle \frac{(-4)^{7}\times (-6)^2 \times 3^{-7}}{(-3)^{5} \times 4^{-11} \times 6^{-3}} \\
  & = \dfrac{(-1)^9 \times (2^2)^7 \times (3 \times 2)^2 \times 3^{-7}}{(-1)^5 \times 3^5 \times (2^2)^{-11} \times (3 \times 2)^{-3}} \\
  & = \dfrac{(-1)^4 \times 2^{14} \times 3^2 \times 2^2  \times 3^{-7}}{3^5 \times 2^{-22} \times 3^{-3} \times 2^{-3}} \\
  & = \boxed{2^{41} \times 3^{-7} }
}

\end{exo}

\begin{exo}[Calculs]
Simplifie chaque expression au maximum.

\reponset{ 
A & = \displaystyle \frac{x^2 +2x +1}{(1+x)^2} \\
  & = \dfrac{(x+1)^2}{(1+x)^2} \\
  & = \boxed{1}
}{
B & = \displaystyle \frac{x^4 - 9}{(x^2+3)^2} \\
  & = \dfrac{(x^2 - 3)(x^2 +3)}{(x^2 +3)^2} \\
  & =\boxed{ \dfrac{x^2 - 3}{x^2 +3}}
}{
C & = \displaystyle \frac{(x-1)(x^2-2x+1)}{(x-1)^{-2}} \\
  & = \dfrac{(x-1)(x-1)^2}{(x-1)^{-2}} \\
  & = \boxed{(x-1)^5}
}
\end{exo}


\end{document}
