Question

Aratati ca urmatoarele propoziti sunt adevarate:
a)6+2√5=(1+√5)²
b)32-10√7=(5-√7)²
c) √16-6√7(tot sub radical)=3-√7
d)√10-2√21(tot sub radical)=√7-√3
e)√6-2√5(tot sub radical)=I1-√5I
Va rog rezolvarile complete.

Answer 1

a) (1+V5)^2=1^2+2V5+(V5)^2=1+5+2V5=6+2V5
b) (5-V7)^2=5^2-2*5*V7+(V7)^2=25-10V7+7=32-10V7
c) Se foloseste formula Va-Vb=V(a+c)/2- V(a-c)/2 unde c=Va^2-b
in cazul nostru V16-6V7=V16-V6^2*7=V16-V252 
c=V16^2-252=V256-252=V4=2
V16-6V7=V(16+2)/2-V16-2/2=V9-V7=3-V7

d) se foloseste aceeasi formula
V10-2V21=V10-V4*21=V10-V84
c=Va^2-b=V10^2-84=V100-84=V16=4
V10-2V21=V(10+4)/2-V(10-4)/2=V7-V3

e) V6-2V5=V6-V4*5=V6-V20
c=V6^2-20=V36-20=V16=4
V6-2V5=V(6+4)/2-V(6-4)/2=V5-1=|1-V5|