题目

几道初一的整式运算题目,会做的来~
1.(a-b)^m(a-b)[2(b-a)]^3[(b-a)^5]^2/(a-b)^m
2.(2006^2+2004^2+```+2^2)-(2005^2+2003^2+```+1)
3.先化简,后求值.
(1)x(x+4)(x-4)-(x+3)(x^2-6x+9)+5x^3y^2/(x^2y^2),其中x=-3.
(2)y(x+y)+(x+y)(x-y)-x^2,其中x=-2,y=1/2.
x^2就表明是2的2次方

提问时间：2021-01-08

答案

1.(a-b)^m(a-b)[2(b-a)]^3[(b-a)^5]^2/(a-b)^m
=(a-b)[2(b-a)]^3[(b-a)^5]^2
=(a-b)*8(b-a)^3*(b-a)^10
=-(a-b)*8(a-b)^3*(a-b)^10
=-8(a-b)^14
2.(2006^2+2004^2+```+2^2)-(2005^2+2003^2+```+1)
=2006^2-2005^2+2004^2-2003^2+```+2^2-1
=(2006+2005)(2006-2005)+(2004+2003)(2004-2003)+```+(2+1)(2-1)
=2006+2005+2004+2003+```+2+1
=(2006+1)*2006/2
=2013021
3.先化简,后求值.
(1)x(x+4)(x-4)-(x+3)(x^2-6x+9)+5x^3y^2/(x^2y^2)
=x(x^2-16)-(x^3-3x^2-9x+27)+5x
=x^3-16x-x^3+3x^2+9x-27+5x
=3x^2-2x-27
=3*9+2*3-27
=6
(2)y(x+y)+(x+y)(x-y)-x^2
=xy+y^2+x^2-y^2-x^2
=xy
=-2*1/2
=-1