AIDEZ-MOI S'IL VOUS PLAIT, URGENT.

Salut,

Exercice 1:
1)
[tex]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} [/tex]
2)
[tex]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} \\ [/tex]
[tex]G = 10\sqrt{3} + 12\sqrt{3} - 10\sqrt{3} \\ G = 12\sqrt{3} [/tex]

Exercice 2:

1)
A = [tex]A = ( \sqrt{7} + \sqrt{5} )( \sqrt{7} - \sqrt{5} ) = \sqrt{7}^{2} -\sqrt{5}^{2} = 7 - 5 = 2\\ [/tex]
[tex]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[/tex]

2)

[tex]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} [/tex]

Exercice 3:

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

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

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

P(abc) = AB + AC + BC = [tex] 4\sqrt{5} + \sqrt{125} + \sqrt{45} = 4\sqrt{5} +5\sqrt{5} + 3\sqrt{5} = 12\sqrt{5}[/tex]
A(abc) = [tex] \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[/tex]

Bonne soirée !