Răspuns :
I5-√5I
5-√5>0
I5-√5I=5-√5
I2-√5I
2-√5<0
I2-√5I=-(2-√5)=-2+√5
a=I5-√5I+I2-√5I-2=5-√5-2+√5-2=5-4=1 deci a e numar natural
fie a= V2-1 ,b=V2+1.Aratati ca 1/a +1/b apartine (2,3)
1/(√2-1)+1/(√2+1)=
=(√2+1+√2-1)/(√2+1)(√2-1)=
=2√2/(2-1)=2√2
2<2√2<3
5-√5>0
I5-√5I=5-√5
I2-√5I
2-√5<0
I2-√5I=-(2-√5)=-2+√5
a=I5-√5I+I2-√5I-2=5-√5-2+√5-2=5-4=1 deci a e numar natural
fie a= V2-1 ,b=V2+1.Aratati ca 1/a +1/b apartine (2,3)
1/(√2-1)+1/(√2+1)=
=(√2+1+√2-1)/(√2+1)(√2-1)=
=2√2/(2-1)=2√2
2<2√2<3
I5-√5I
5-√5>0
I5-√8l=5-√6
I2-√5I
2-√5.0
I2-√5I=-(2-√5)=-2+√5
a=9l
-√5I+I2-√5I-2=5-√5-2+√5-2=5-4=1
5-√5>0
I5-√8l=5-√6
I2-√5I
2-√5.0
I2-√5I=-(2-√5)=-2+√5
a=9l
-√5I+I2-√5I-2=5-√5-2+√5-2=5-4=1