若(x²+y²)(x²-1+y²)-12=0,则x²+y²=
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若(x²+y²)(x²-1+y²)-12=0,则x²+y²=
若(x²+y²)(x²-1+y²)-12=0,则x²+y²=
若(x²+y²)(x²-1+y²)-12=0,则x²+y²=
令x²+y²=z,则可知z大于等于0.原式可化为z(z-1)-12=0 解一元二次方程,化为(z+3)(z-4)=0 则z=-3或z=4.又知z大于等于0,则排除-3得z=4,即x²+y²=4.
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