设X^2+2X-1=0.Y^4-2Y^2-1=0,且xy^2不等于1,求[(xy^2+y^2+1)/x]^1999的值

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设X^2+2X-1=0.Y^4-2Y^2-1=0,且xy^2不等于1,求[(xy^2+y^2+1)/x]^1999的值设X^2+2X-1=0.Y^4-2Y^2-1=0,且xy^2不等于1,求[(xy^

设X^2+2X-1=0.Y^4-2Y^2-1=0,且xy^2不等于1,求[(xy^2+y^2+1)/x]^1999的值
设X^2+2X-1=0.Y^4-2Y^2-1=0,且xy^2不等于1,求[(xy^2+y^2+1)/x]^1999的值

设X^2+2X-1=0.Y^4-2Y^2-1=0,且xy^2不等于1,求[(xy^2+y^2+1)/x]^1999的值
X^2+2X-1=0
1+2/x-1/x^2=0
则1/x^2-2/x-1=0
同时:Y^4-2Y^2-1=0
由于1/x≠y^2
故1/x,y^2为方程a^2-2a-1=0的两根
由韦达定理:故有
1/x+y^2=2
y^2/x=-1
[(xy^2+y^2+1)/x]^1999
=[1/x+y^2/x+y^2]^1999
=1^1999=1