分式(分母）解

因为x^2-y^2=(x+y)(x-y)…………这一步是利用平方差公式
x^2+xy=x(x+y) 这一步是提取公因式
所以1/(x^2-y^2)与1/（x^2+xy)的最简分母为x(x+y)(x-y)
1/（x^2-y^2)=1/(x+y)(x-y)=x/x(x+y)(x-y)
1/(x^2+xy)=1/x(x+y)=(x-y)/x(x+y)(x-y)

因为x^2-y^2=(x+y)(x-y),
x^2+xy=x(x+y),
所以1/(x^2-y^2)与1/（x^2+xy)的最简分母为x(x+y)(x-y),
1/（x^2-y^2)=x/[x(x+y)(x-y)],
1/(x^2+xy)=(x-y)/[x(x+y)(x-y)].

因为 x^2-y^2=(x+y)(x-y)
x^2+xy=(x+y)x
所以1/(x^2-y^2)与1/（x^2+xy)的最简分母为(x+y)
1/（x^2-y^2)=1/(x+y)(x-y)
1/(x^2+xy)=1/(x+y)x