Question

As vrea sa ma ajutati cu un exercitiu la matematica clasa a 10-a

Answer 1

$\sqrt{17-4 \sqrt{9+4 \sqrt{5} } } = \sqrt{5} -2$    | ridicam la patrat
$17-4 \sqrt{9+4 \sqrt{5} } =9-4 \sqrt{5}$
$8-4 \sqrt{9+ \sqrt{80} } =-4 \sqrt{5}$   impartim cu 4
$2- \sqrt{9+ \sqrt{80} } =- \sqrt{5}$   
Folosim formula radicalilor  compusi!!! ______________
$\sqrt{a+ \sqrt{b} } = \sqrt{ \frac{a+ \sqrt{a^2-b} }{2} } +\sqrt{ \frac{a- \sqrt{a^2-b} }{2} }$

$\sqrt{9+ \sqrt{80} } = \sqrt{ \frac{9+ \sqrt{81-80} }{2} } +\sqrt{ \frac{9- \sqrt{81-80} }{2} }$

$\sqrt{5} + \sqrt{4}$
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$2- \sqrt{5} - \sqrt{4} =- \sqrt{5}$
Se reduce tot , 0=0(A) 

$b) \frac{1}{ \sqrt{2} +1}+...+ \frac{1}{ \sqrt{100}+ \sqrt{99} } =9$
Amplificam cu conjugata:
$\frac{ \sqrt{2}-1 }{2-1} + \frac{ \sqrt{3}- \sqrt{2} }{3-2} +...+ \frac{ \sqrt{100}- \sqrt{99} }{100-99}$
$\sqrt{2} -1+ \sqrt{3}- \sqrt{2} +...+ \sqrt{99}- \sqrt{98} + \sqrt{100} - \sqrt{99}$
Se duce tot, inafara de -1 si 10 , care adunate dau exact 9.

c) proprietatea log:  b(log in baza b din x) =x
36 la (log baza 6 din 5)=6 la(log baza 6 din 5) ori 6 la( log baza 6 din 5)
=5*5=25. (uita-te in atasament sau http://prntscr.com/5a3v50