Question

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

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Answer 1

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