Question

2001ⅹ2003/2002+2002ⅹ2004/2003+2002x4005/2003

Answer 1

原式=(2002-1)*(2002+1)/2002+(2003-1)*(2003+1)/2003+4005/(2002*2003)
=(2002^2-1)/2002+(2003^2-1)/2003+4005/(2002*2003)
=2002-1/2002+2003-1/2003)+4005/(2002*2003)
=2002+2003-(1/2002+1/2003)+4005/(2002*2003)
=4005-(2003+2002)/(2002*2003)+4005/(2002*2003)
=4005

Answer 2

2001ⅹ2003/2002+2002ⅹ2004/2003+2002x4005/2003
设 A=2002
原式
=(A-1)(A+1)/A +A(A+2)/(A+1) +A(2A+1)/(A+1)
=(A²-1)/A +[(A²+2A)+(2A²+A)]/(A+1)
= (A²-1)/A +(3A²+3A)/(A+1)
= (A²-1)/A +3A(A+1)/(A+1)

= (A²-1)/A +3A²/A
=(A²-1+3A²)/A
=[(2A)²-1]/A
=(2A+1)(2A-1)/A
=[(2x2002+1)(2x2002-1)]/2002
=[(2x2002)²+2x2002-2x2002-1]/2002
=[(2x2002)²-1]/2002
= (2x2002)²/2002 -1/2002
=8008 -1/2002
=8007又2001/2002