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\begin{document}%\everymath{\displaystyle}
\boxit{4pt}{Minitest 10}\rightarrowfill\quad\boxit{3pt}{Corrigé}\bigskip

\doublespacing

Calculer : 
$5! ={\red 5\times 4\times 3\times 2\times 1 = 120}$

\qquad 

$\dfrac{7\times 4!}{3!\times 5!} = {\red\dfrac{7\times 4!}{3\times2\times5\times 4!} = \dfrac{7}{30}}$

\vskip 0.5cm

Exprimer le plus simplement possible : 

$\ds\binom{n}{2} = {\red \dfrac{n!}{(n-2)!2!} = \dfrac{n(n-1)}{2}}$


\bigskip

$\ds\binom{n+1}{n-1} = {\red \dfrac{()n-1)!}{()(n+1)-(n-1))!(n-1)!} = \dfrac{(n+1)(n)}{2}}$


\vskip 1cm

Soit $A(2;3)$, $B(-1;5)$ et $C(7;-6)$

Calculer les valeurs exactes de $AB$ et de $AC$

{\red $AB = \sqrt{(-1-2)^2+(5-3)^2} = \sqrt{9+4} = \sqrt{13}$

$AC = \sqrt{(7-2)^2+(-6-3)^2} = \sqrt{25+81} = \sqrt{106}$
}

Déterminer les coordonnées du milieu $I$ de $[AB]$

{\red $I\left(\frac12;4\right)$}

Calculer les composantes de $\vect{AB}$ et de $\vect{AC}$

{\red $\vect{AB}\begin{pmatrix}-3\cr 2\end{pmatrix}$\qquad $\vect{AC}\begin{pmatrix}5\cr-9\end{pmatrix}$}

Calculer $\vect{AB}\cdot\vect{AC}$

{\red $\vect{AB}\cdot\vect{AC} = -15-18 = -33$}

Déterminer la valeur exacte de $\cos(\vect{AB},\vect{AC})$ puis la valeur au dixième de degré près de l'angle $(\vect{AB},\vect{AC})$

{\red $\cos(\vect{AB},\vect{AC}) = \dfrac{-33}{\sqrt{13}\times\sqrt{106}}$ donc la clculatrice donne $(\vect{AB},\vect{AC})\approx152,7^o$ }

\end{document} 