Question

2002的2次方+2002-2003的2次方等于多少

2002的2次方+2002-2003的2次方等于多少 步骤 证明81的7次方-27的9次方-9的13次方能被45整除

Answer 1

2002的2次方+2002-2003的2次方

=(2002+2003)(2002-2003)+2002

=-2003

81^7-27^9-9^13

=3^28-3^27-3^26

=3^26(9-3-1)

=3^24*45

所以能被45整除

Answer 2

2002的2次方+2002-2003的2次方
=(2002+2003)(2002-2003)+2002
=-4005+2002
=-2003

81的7次方-27的9次方-9的13次方
=3^28-3^27-3^26
=3^26(9-3-1)
=3^26*5
=45*3^24能被45整除

Answer 3

2002^2+2002-2003^2
=(2002+1)2002-2003*2003
=2003(2002-2003)
=-2003
81^7-27^9-9^13
=9^14-3*9^13-9^13
=9^13(9-3-1)
=5*9^13
=45*9^12

Answer 4

1,
2002^2+2002-2003^2
=2002^2+2002-(2002+1)^2
=2002^2+2002-2002^2-2*2002-1
=-2003

2,
81^7-27^9-9^13
=3^28-3^27-3^26
=3^26(9-3-1)
=3^2*3^24*5
=45*3^24