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Meilleure réponse !
  • Omnes
  • Modérateur confirmé
2014-02-18T20:15:14+01:00
Salut,

Exercice 1:
1)
A =  \sqrt{2} *  \sqrt{8}   =  \sqrt{2*8}  =  \sqrt{16} = 4\\
B=  \sqrt{10} *  \sqrt{1000} =  \sqrt{10*1000} =  \sqrt{10000} =  100\\
C =  \frac{ \sqrt{3} }{\sqrt{27}} = \frac{ \sqrt{3} *   \sqrt{1} }{ \sqrt{3} * \sqrt{9}}=  \frac{ \sqrt{1} }{ \sqrt{9} }  =  \frac{1}{3}
2)
D =  \sqrt{8} =  \sqrt{4*2} =  \sqrt{4} *  \sqrt{2} = 2 \sqrt{2} \\
E =  5\sqrt{6} * 2 \sqrt{3} =  5*2\sqrt{6*3} \\ 
E = 10 \sqrt{18} = 10 \sqrt{9*2} = 10 \sqrt{9} *  \sqrt{2} = 10 * 3 \sqrt{2}  = 30 \sqrt{2}\\
F =  \sqrt{20} + \sqrt{125} - \sqrt{45} +  \sqrt{5} \\
F = \sqrt{4*5} +  \sqrt{25*5} -  \sqrt{9*5} + \sqrt{5} \\
F = 2\sqrt{5} + 5\sqrt{5}  - 3\sqrt{5} +\sqrt{5} \\
F = 5\sqrt{5} \\
G = 5 \sqrt{12} + 3  \sqrt{48} - 2 \sqrt{75} \\
G = 5 \sqrt{4*3} +  3\sqrt{16*3} - 2 \sqrt{25*3}  \\

G = 10\sqrt{3} + 12\sqrt{3} - 10\sqrt{3} \\
G = 12\sqrt{3}

Exercice 2:

1)
A = A = ( \sqrt{7} +  \sqrt{5} )( \sqrt{7} -  \sqrt{5} ) =  \sqrt{7}^{2} -\sqrt{5}^{2} = 7 - 5 = 2\\


B =  ( 2\sqrt{2} - 2 )( 3\sqrt{2} +3 ) = 3\sqrt{2} * 2\sqrt{2} + +3*2\sqrt{2} - 2*3\sqrt{2} - 2*3\\
B = 6*2 + 5\sqrt{2} - 5\sqrt{2} - 6 
B = 12 - 6
B = 6

2)

C = ( \sqrt{2} -  \sqrt{5})^{2} = \sqrt{2}^{2} - 2*\sqrt{2}* \sqrt{5} + \sqrt{5}^{2} = 2 - 2 \sqrt{5*2} + 5 \\
C = 7 -  2\sqrt{10}

Exercice 3:

On a AB = 4 \sqrt{5}; AC =  \sqrt{125} ;BC =  \sqrt{45}
On compare AC² et AB² + BC²

AC² = 125
AB² + BC² = (4 \sqrt{5})^{2}  +  \sqrt{45}^{2} = 16*5 + 45 = 80 + 45 = 125

On a : AC² = AB² + BC² donc le triangle ABC est rectangle selon la reciproque du théorème de Pythagore.

P(abc) = AB + AC + BC =  4\sqrt{5} +  \sqrt{125} +  \sqrt{45}  =  4\sqrt{5} +5\sqrt{5} + 3\sqrt{5} = 12\sqrt{5}
A(abc) =  \frac{AB*AC}{2} =  \frac{4 \sqrt{5} *  \sqrt{125} }{2} =  \frac{4* \sqrt{5*125} }{2} =  \frac{4* \sqrt{625} }{2} =  \frac{100}{2} = 50

Bonne soirée !