(1)若
+(b-2)2=0,A=3a2-6ab+b2,B=-a2-5,求A-B的值.
(2)试说明:无论x
,y取何值时,代数式
(x3+3x2y-5xy+6y3)+(y3+2xy2+x2y-2x3)-(4x2y-x3-3xy2+7y3)的值是常数.
1.(1)∵A=3a2-6ab+b2,B=-a2-5,∴A-B=(3a2-6ab+b2)-(-a2
-5)=4a2-6ab+b2+5.
又∵
+(b-2)2=0,∴A-B=4×12-6×1×2+22+5=1.
(2)原式化简值结果不含x,y字母,即原式=0.∴无论x,y取何值,原式的值均为常数0.
略