1．当x=0.3,y=0.7时,求分式x^3-x^2-xy^2+y^3/-x^4+2x^2y^2-y^4的值.2．当a=3/2,b_数学解答

1．当x=0.3,y=0.7时,求分式x^3-x^2-xy^2+y^3/-x^4+2x^2y^2-y^4的值.2．当a=3/2,b=-1/2时,求代数式(a-b+4b/a-b)*(a+b-4ab/a+b)的值.求求了!

人气：244 ℃　时间：2020-06-13 16:07:28

解答

x^3-x^2-xy^2+y^3/-x^4+2x^2y^2-y^4
=[(x^3+y^3)-xy(x+y)]/[-(x^2-y^2)^2]
=[(x+y)(x^2-xy+y^2)-xy(x+y)]/[-(x^2-y^2)^2]
=[x^2-2xy+y^2]/[-(x^2-y^2)^2]
=-(x-y)^2/(x-y)^2(x+y)^=-1/(x+y)^2=-1
A-B=2,A+B=1,AB=-3/4,直接带入求值就可以了