Question

已知3x²+xy-2y²=0(x≠0,y≠0),求x/y-y/x-x²+y²/xy的值

Answer 1

3x²+xy-2y²=0(x≠0,y≠0),

3+y/x-2y^2/x^2=0,假设y/x=t

2t^2-t-3=0

(2t-3)(t+1)=0

t=3/2,t=-1

x/y-y/x-x²+y²/xy

(x^2-y^2-x^2-y^2)/(xy)

=-2y/x

=-2t

所以：

x/y-y/x-x²+y²/xy的值 3或2