Question

Fie fractia ordinara 6n+20 supra 2n+1 unde n este un numar natural oarecare.Scoateti intregii din fractie. a) daca n=8 b)daca n apartine N \{1,2,3,...,8}

Answer 1

a)
(6x8+20)/(2x8+1)=
=68/17=4 intregi

b)
n=N-{1,2,3,...,8}, adica n>8

n>8  l  x2
2n>16  l +1
2n+1>17  l inversam
1/(2n+1)<1/17  l  x17

17/(2n+1)<1
 

(6n+20)/(2n+1)=(6n+3+17)/(2n+1)=(6n+3)/(2n+1)+17/(2n+1)=

=6(2n+1)/(2n+1)+17/(2n+1)=6+17/(2n+1)

Partea intreaga este 6 si partea fractionara 17/(2n+1)<1 (am demonstrat)