题目

问大家一道初一正式乘法题!急呃~
(x+2y+3)²-(x-2y+3)(x-2y-3)
原式=
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一定要对啊，要考的，一道6分呢……

提问时间：2021-10-27

答案

(x+2y+3)²-(x-2y+3)(x-2y-3)
=(x+2y+3)²-((x-2y)+3)((x-2y)-3)
=((x+2y)+3)²-(x-2y)^2+9
=(x+2y)^2+6(x+2y)+9-(x-2y)^2+9
=(x+2y)^2-(x-2y)^2+6(x+2y)+18
=(x+2y+x-2y)(x+2y-x+2y)+6(x+2y)+18
=2x*4y+6x+12y+18
=8xy+6x+12y+18
=2xy+6x+4xy+12y+18
=2x(y+3)+4y(x+3)+18