给出三个多项式2x2+3xy+y2,3x2+3xy,x2+xy,请你任选两个进行加(或减)法运算,再将结果分解因式.
解:(2x2+3xy+y2)-(3x2+3xy)
=2x2+3xy+y2-3x2-3xy
=y2-x2
=(y+x)(y-x);
(2x2+3xy+y2)+(3x2+3xy)
=5x2+6xy+y2
=(5x+y)(x+y);
(2x2+3xy+y2)+(x2+xy)
=3x2+4xy+y2
=(3x+y)(x+y);
(2x2+3xy+y2)-(x2+xy)
=x2+2xy+y2
=(x+y)2;
(3x2+3xy)+(x2+xy)
=4x2+4xy
=4x(x+y);
(3x2+3xy)-(x2+xy)
=2x2+2xy
=2x(x+y);
分析:将任选两个进行加(或减)法运算,求得结果分解因式即可;注意答案不唯一.
点评:考查整式的加减与分解因式.本题比较简单.