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Simplificati fractiile astfel incat sa devina ireductibile:
A) 3 la 2003+ 3 la 2005 supra 3 la 2007- 3 la 2005
B) 2 la 2007- 6×2 la 2002 supra 2 la 2001+ 2 la 2003+ 2 la 2004 .
URGENT!!!! MULTUMESC!

Răspuns :

Renatemambouko
A) (3^2003+ 3^2005)/ (3^2007- 3^2005)=
=3^2003(1+ 3^2)/ 3^2005(3^2- 1)=
=3^2003(1+ 9)/ 3^2005(9- 1)=
=10/(3^2×8)=
=10/72=
=5/36

B) (2^2007- 6×2^2002)/ (2^2001+ 2^2003+ 2^2004)=
= (2^2007- 3×2^2003)/ 2^2001(1+ 2^2+ 2^3)=
= 2^2003(2^4- 3)/ 2^2001(1+ 2^2+ 2^3)=
= 2^2003×13)/ 2^2001×13=
=2^2=4