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Symy
a fost răspuns

Cum se rezolvă : 1*2+2*3+3*4+...+2013*2014 supra 2*4+4*6+6*8+...+4026*4028 = 1 supra x la puterea a 2 -a   ??? 

Răspuns :

Andrarac
1*2+2*3+...+2013*2014=t
[tex] \frac{t}{4t}= \frac{1}{4} [/tex]
=>[tex] x^{2} [/tex]=4
=>x=2
Georgynico
[tex][ \frac{1*2+2*3+....+2013*2014 }{2*4+4*6+...+4026*4028} = \frac{1*2+2*3+....+2013*2014}{2*(1*2+2*3+....+2013*2014)}= \frac{1}{2} = \frac{1}{x patrat} [/tex]
de unde x patrat e 2
adik este + sau -[tex] \sqrt{2} [/tex]