Question

comparati numerele a=3la puterea2000 -3la puterea 1999 -3 la puterea 1997 si b= 2la puterea 2002 -2 la puterea 2001 +2 la puterea 1997

Answer 1

a=3^2000 -3^1999 -3^1997 =
=3^1997× (3^3-3^2-3^0)=3^1997×(27-9-1)=3^1997 ×17
b= 2^2002 -2^2001 +2^1997 =
=2^1997×(2^5-2^4+2^0)=2^1997×(32-16+1)=2=2^1997×17
3>2
rezulta ca  a>b