Réponses

  • Omnes
  • Modérateur confirmé
2014-10-30T22:51:36+01:00
Salut,

Exercice 27:

 \frac{2}{ \sqrt{3} } =  \frac{2* \sqrt{3} }{\sqrt{3}*\sqrt{3}} =  \frac{2 \sqrt{3}}{3}
 \frac{7}{2 \sqrt{5} }  =  \frac{7* \sqrt{5} }{2 \sqrt{5}* \sqrt{5}} =  \frac{7\sqrt{5}}{2*5} =  \frac{7\sqrt{5}}{10}
 \frac{ \sqrt{3} }{4 \sqrt{2} } = \frac{ \sqrt{3} * \sqrt{2} }{4\sqrt{2} * \sqrt{2} } = \frac{ \sqrt{3*2} }{4*2} = \frac{ \sqrt{6} }{8}
 \frac{3 \sqrt{3} }{ \sqrt{8} } =  \frac{3 \sqrt{3} *  \sqrt{8}  }{    \sqrt{8}*  \sqrt{8}} =  \frac{3 \sqrt{4*2*3} }{8} =  \frac{3 \sqrt{24} }{8} =  \frac{3* \sqrt{4*6} }{8}  =  \frac{3*2 \sqrt{6} }{8} =  \frac{6 \sqrt{6} }{8} =  \frac{3 \sqrt{6} }{4}


Exercice 28

 \frac{-1}{ \sqrt{2} } =  \frac{-1* \sqrt{2} }{\sqrt{2}*\sqrt{2}} =  \frac{- \sqrt{2} }{2} \\

 \frac{-4 \sqrt{3} }{ \sqrt{6} } = \frac{-4\sqrt{3} *  \sqrt{6}}{\sqrt{6}*\sqrt{6}} =  \frac{-4* \sqrt{3*6} }{6} =  \frac{-4 \sqrt{9*2} }{6} =  \frac{-4* 3* \sqrt{2} }{6} =  \frac{-12\sqrt{2} }{6}= -2 \sqrt{2}

 \frac{2 \sqrt{6} }{3 \sqrt{8} } =  \frac{2 \sqrt{6}*\sqrt{8}}{3\sqrt{8}*\sqrt{8}} =  \frac{2 \sqrt{48} }{24} =  \frac{ \sqrt{16*3} }{12} =  \frac{4 \sqrt{3} }{12} =  \frac{ \sqrt{3} }{3}

 \frac{ \sqrt{7} *  \sqrt{6} }{ \sqrt{2} *  \sqrt{3}  } =  \frac{\sqrt{7} *  \sqrt{6}}{ \sqrt{2*3} } =  \frac{\sqrt{7} *  \sqrt{6}}{ \sqrt{6} } =  \sqrt{7}


Exercice 29

D =  \frac{5 \sqrt{12} }{2 \sqrt{3} } =  \frac{5 \sqrt{4*3} }{2 \sqrt{3}} =  \frac{5*2\sqrt{3}}{2\sqrt{3}} =  \frac{10\sqrt{3}}{2\sqrt{3}} = 5

D = 5, donc D est un nombre entier.

Bonne soirée !