Question

Se considera numerele a=√3^2+4^2+(√8-√2)^2 si b=10-(√27-√3)^2. Arata ca a>b.
Ajutati-ma va rog dau 35 puncte.

Answer 1

a=3+16+(8+2-2√2*8)=19+(10-2√16)=19+(10-8)=19+2=21
b=10-(27+3-2√27*3)=10-(30-2√81)=10-(30-18)=10-12=-2
 21>-2
a>b

Answer 2

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