Question

请写出计算的具体过程

Answer 1

Answer 2

=(2004²-2004＋1)/(2004²-2004×2003+2003²)
=(2004²-2x2004＋1+2004)/(2004²-2x2004×2003+2003²+2004x2003)
=[(2004-1)²+2004]/[(2004-2003)²+2004x2003]
=[2003²+2004]/[1²+(2003+1)x2003]
=[2003²+(2003+1)]/[1+2003²+2003]
=[2003²+2003+1]/[2003²+2003+1]
=1

Answer 3

2004*2004-2004+1/2004*2004-2004*2003+2003*2003
=2004*（2004-1）+1/2004*（2004-2003）+2003*2003
=2004*2003+1/2004+2003*2003
=2004*2003+1/2003+1+2003*2003
=2004*2003+1/2003*（2003+1）+1
=2004*2003+1/2004*2003+1
=1

Answer 4