矩形面积=9(x+y)^2-y^2=(3x+3y)^2-y^2
=(3x+3y+y)(3x+3y-y)
=(3x+4y)(3x+2y) 又知(X>0,Y>0,且都为整数),所以3x+4y为长,3x+2y为宽,
所以该矩形周长的代数式为
2(3x+4y+3x+2y)
=2(6x+6y)
=12x+12y详细吗