问几道初中代数题

1.分解因式
81(x+y)^4-16(x-y)^4
2.已知矩形的面积是x^2-2x-35(x>7),其中一边长是x-7，求表示矩形的另一边的代数式。
3.已知正方形的面积是9x^2+6xy+y^2(x>0,y>0),求表示该正方形的边长的代数式。

81(x+y)^4-16(x-y)^4
=(9(x+y)^2+4(x-y)^2)(9(x+y)^2-4(x-y)^2)
=(9(x+y)^2+4(x-y)^2)(3(x+y)+2(x-y))(3(x+y)-2(x-y))
=(13x^2+10xy+13y^2)(5x+y)(x+5y)

x^2-2x-35=(x-7)(x+5)
所以另一边是x+5

9x^2+6xy+y^2=(3x+y)^2
3x+y>0
所以边长是3x+y

1.分解因式
81(x+y)^4-16(x-y)^4
=[9(x+y)^2-4(x-y)^2][9(x+y)^2+4(x-y)^2]
=[3(x+y)-2(x-y)][3(x+y)+2(x-y)][9(x+y)^2+4(x-y)^2]
=[(3x+3y)-(2x-2y)][(3x+3y)+(2x-2y)][9(x+y)^2+4(x-y)^2]
=(3x+3y-2x+2y)(3x+3y+2x-2y)[9(x+y)^2+4(x-y)^2]
=(x+5y)(5x+y)[9(x+y)^2+4(x-y)^2]
=(x+5y)(5x+y)[9(x^2+y^2+2xy)+4(x^2+y^2-2xy)]
=(x+5y)(5x+y)[(9x^2+9y^2+18xy)+(4x^2+4y^2-8xy)]
=(x+5y)(5x+y)(13x^2+13y^2+10xy)

2.
因为x^2-2x-35=(x-7)(x+5),所以矩形的另一边的代数式是x+5。

3.
因为9x^2+6xy+y^2=(3x+y)^2,所以正方形的边长的代数式是3x+y。

1.
=(9(x+y)²+4(x-y)²)(9(x+y)²-4(x-y)²)
=(13x²+10xy+13y²)(3x+3y+2x-2y)(3x+3y-2x+2y)
=(13x²+10xy+13y²)(5x+y)(x+5y)