已知x²+4y²-4x+4y+5=0,求:
3个回答

x^2+4y^2-4x+4y+5=0.

(x-2)^2-4+4*(y+1/2)^2-1+5=0.

(x-2)^2+4(y+1/2)^2=0.

(x-2)^2=0,x=2;

(y+1/2)^2=0,y=-1/2.

原式=[(x-y)(x+y)((X^2+Y^2)]/[(2x-y)(x+y)]*(2x-y)/y(x-y)/[x^2+y^2)^2/y^2].分子、分母约去公因子后,得:

原式=y/(x^2+y^2).(1)

将x=2,y=-1/2代人(1),得:

原式=(-1/2)/[(2^2+(-1/2)^2].

=--2/17.

∴(x^4-y^4)/[(2x-y)(x+y)]*(2x-y)/(xy-y^2)/[(x^2+y^2)^2/y^2]=-2/17 .