Question

Exercitiul 2, testul 2
Va rog!!

Answer 1

[a;b]∩Z=∅⇒a<b<a+1
atunci |b-a-1|=a+1-b

a+1-b=a²+b²+a/2-2b+21/16

a²-a/2+b²-b+5/16=0

a²-a/2+1/16+b²-b+1/4=0
(a-1/4)² +(b-1/2)²=0
cum c²≥0, ∀c∈R
relatia poate fi valabila numai pt.
a-1/4 =0
si
b-1/2=0

deci a=1/4 si b=1/2
 as tricky as that!!!

verificare
[1/4;1/2]∩Z=∅, adevarat

|1/2-1/4-1|=1/16+1/4+1/8-1+21/16
|-3/4|=5/16+2/16-16/16+21/16
3/4=(28-16)/16
3/4=12/16
3/4=3/4
adevarat
adevarate ambele, bine rezolvat