Exercice 1 :
1) A = [tex] \sqrt{700} - 8\sqrt{28} + 4\sqrt{7} [/tex]
= [tex] \sqrt{7(100)} - 8\sqrt{7(4)} + 4\sqrt{7} [/tex]
= [tex] \sqrt{100} × \sqrt{7} - 8 \sqrt{4} × \sqrt{7} + 4\sqrt{7} [/tex]
= [tex] 10\sqrt{7} - 16\sqrt{7} + 4\sqrt{7} [/tex]
= [tex] \sqrt{7} ( 10 - 16 + 4 ) [/tex]
A = [tex] -2\sqrt{7} [/tex]
B = [tex] 3\sqrt{45} + 2\sqrt{20} - 4 \sqrt{80} [/tex]
= [tex] 3\sqrt{9(5)} + 2\sqrt{4(5)} - 4 \sqrt{16(5)} [/tex]
= [tex] 9\sqrt{5} + 4\sqrt{5} - 16\sqrt{5} [/tex]
= [tex] \sqrt{5} ( 9 + 4 - 16 ) [/tex]
B = [tex] -3\sqrt{5} [/tex]
C = [tex]3 \sqrt{250} - 2 \sqrt{810} + \sqrt{1210} [/tex]
= [tex]3 \sqrt{25(10)} - 2 \sqrt{81(10)} + \sqrt{121(10)} [/tex]
= [tex] 15\sqrt{10} - 18\sqrt{10} + 11\sqrt{10} [/tex]
= [tex] \sqrt{10} ( 15 - 18 + 11 )[/tex]
C = [tex] 8\sqrt{10} [/tex]
D = [tex](2,6 \sqrt{42} ) (5\sqrt{18}) \sqrt{28} [/tex]
= [tex]13 \sqrt{42(18)(28)} [/tex]
= [tex]13 \sqrt{42(9)(2)(4)(7)} [/tex]
= [tex]13(3)(2) \sqrt{42(2)(7)} [/tex]
= [tex]13(3)(2)(14) \sqrt{3} [/tex]
D = [tex]1092 \sqrt{3} [/tex]
2) E = [tex] (7 - 6\sqrt{3} )^{2} [/tex]
= 49 - 84√3 + 108
E = 157 - 84√3
F = [tex] \sqrt{ \frac{72}{75} } (\frac{5}{ \sqrt{48}}) [/tex]
= [tex] \frac{\sqrt{72}}{\sqrt{75}} (\frac{5}{ \sqrt{48}}) [/tex]
= [tex] \frac{3\sqrt{8}}{5\sqrt{3}} (\frac{5}{ 4\sqrt{3}}) [/tex]
= [tex] \frac{15\sqrt{8}}{5\sqrt{3}4\sqrt{3}} [/tex]
= [tex] \frac{15\sqrt{8}}{60} [/tex]
F = [tex] \frac{\sqrt{8}}{4} [/tex]
Exercice 2 :
1) S = DC × AH
AH = S/DC
AH = [tex] \frac{24}{ 4\sqrt{6} } [/tex]
= [tex] \frac{24(4\sqrt{6})}{ 96 } [/tex]
= [tex] \frac{96\sqrt{6}}{ 96 } [/tex]
AH = [tex] \sqrt{6} [/tex]
2) DH = DC - CH
DH = [tex] 4\sqrt{6} - 3\sqrt{6} [/tex]
DH = [tex] \sqrt{6} [/tex]
3) Dans le triangle ABI :
- ( AB ) // ( CH )
- H ∈ ( AI )
- C ∈ ( IB )
Selon le théorème de Thalès nous avons que :
[tex] \frac{IH}{AI} = \frac{CH}{AB} [/tex]
Donc : [tex] IH = \frac{AI(CH)}{AB} [/tex]
IH = [tex] \frac{(IH+AI)(CH)}{AB} [/tex]
IH = [tex] \frac{ 3\sqrt{6} (IH) + 18 }{ 4\sqrt{6}} [/tex]
= [tex] \frac{ 3\sqrt{6} (IH)}{ 4\sqrt{6}} [/tex] + [tex] \frac{18 }{ 4\sqrt{6}} [/tex]
= (IH√6)/6+ (9√6)/2
IH - (IH√6)/6 = (9√6)/2
IH ( 1 - √6/6) = (9√6)/2
IH = [tex] \frac{(9√6)/2}{1 - √6/6} [/tex]
IH = [tex] \frac{27(1+ \sqrt{6}) }{2} [/tex]
Exercice 3 :
Je ne voit pas trop quoi répondre a part un rectangle ...
Exercice 4 :
N = [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{31+ \sqrt{23+1+ \sqrt{8+ \sqrt{1} } } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{31+ \sqrt{23+\sqrt1+ \sqrt{8+1 } } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{31+ \sqrt{23+\sqrt1+ \sqrt{9 } } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{31+ \sqrt{23+\sqrt1+3 } } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{31+ \sqrt{23+\sqrt4 } } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{31+ \sqrt{23+2 } } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{31+ \sqrt{25 } } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{31+ 5 } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ \sqrt{36 } } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{10+ 6 } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+\sqrt{16 } } } [/tex]
= [tex] \sqrt{73+ \sqrt{60+4} } } [/tex]
= [tex] \sqrt{73+ 8 } [/tex]
= [tex] \sqrt{81} [/tex]
N = 9
