2.Fie f, numarul fetelor si b, numarul baietilor
avern f+b=36
b=f+40f/100=f+2f/5=7f/5
in locuim pe b cu 7f/5
7f/5+f=36
12f/5=36 impartim relatia cu 12 in stanga si in dreapta
f/5=3
f=15
rezulta baieti=36-15=21
verificare
2*15/5=6
15+6=21, problema e bine rezolvata
x( 1 +1/2 +1/3+....+1/2015)=1-1/2 +1-2/3+1-3/4+....+1-2014/2015)
x( 1 +1/2 +1/3+....+1/2015)= 1 +1/2 +1/3+....+1/2015)
x= 1 +1/2 +1/3+....+1/2015):( 1 +1/2 +1/3+....+1/2015)
x=1
x= 1+6+6²+6³+.....6^2015= (6^2016-1)/(6-1)= (6^2016-1)/5
5x+1 =(6^2016-1) *5/5 +1= 6^2016-1=1= 6^2016
√(5x+1)=√6^2016=√(6^1008)²=6^1008