已知6x^2-xy-15y^2=0(xy≠0),求(x^2-2xy+4y^2)/(2x^2-3xy-3y^2)的值.
1个回答

已知6x^2-xy-15y^2=0(xy≠0)

(2x+3y)(3x-5y)=0

2x+3y=0或3x-5y=0

(1)2x+3y=0

2x=-3y

x/y=-3/2

(x^2-2xy+4y^2)/(2x^2-3xy-3y^2)的值

=y^2((x/y)^2-2x/y+4)/y^2(2(x/y)^2-3x/y-3)

=((x/y)^2-2x/y+4)/(2(x/y)^2-3x/y-3)

=((-3/2)^2-2*(-3/2)+4)/(2(-3/2)^2-3*(-3/2)-3)

=(9/4+3+4)/(9/2+9/2-3)

=(37/4)/6

=37/24

(2)3x-5y=0

3x=5y

x/y=5/3

(x^2-2xy+4y^2)/(2x^2-3xy-3y^2)的值

=y^2((x/y)^2-2x/y+4)/y^2(2(x/y)^2-3x/y-3)

=((x/y)^2-2x/y+4)/(2(x/y)^2-3x/y-3)

=((5/3)^2-2*(5/3)+4)/(2(5/3)^2-3*(5/3)-3)

=(25/9-10/3+4)/(50/9-5-3)

=(31/9)/(-22/9)

=-31/22