x²+3x-2=0.(1)；y²+3y-2=0.(2)
(1)-(2)得x²-y²+3(x-y)=0
分解因式得 (x+y)(x-y)+3(x-y)=0
提公因式得 (x-y)[(x+y)+3]=0
因为x≠y,即x-y≠0,故必有x+y+3=0,即有x+y=-3.(3)
(1)+(2)得x²+y²+3(x+y)-4=x²+y²-9-4=x²+y²-13=0,故x²+y²=13.(4)
将(3)两边平方得x²+y²+2xy=9,故xy=[9-(x²+y²)]/2=(9-13)/2=-2
∴(y/x)+(x/y)=(x²+y²)/xy=-13/2