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=](/uploads/image/z/2559887-71-7.jpg?t=%E8%8B%A5%28x%26%23178%3B%2By%26%23178%3B%29%28x%26%23178%3B-1%2By%26%23178%3B%29-12%3D0%2C%E5%88%99x%26%23178%3B%2By%26%23178%3B%3D)
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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