(x²+y²+1)(x²+y²+3)=8,则x²+y²多少
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![(x²+y²+1)(x²+y²+3)=8,则x²+y²多少](/uploads/image/z/6852806-62-6.jpg?t=%EF%BC%88x%26%23178%3B%2By%26%23178%3B%2B1%EF%BC%89%EF%BC%88x%26%23178%3B%2By%26%23178%3B%2B3%EF%BC%89%3D8%2C%E5%88%99x%26%23178%3B%2By%26%23178%3B%E5%A4%9A%E5%B0%91)
(x²+y²+1)(x²+y²+3)=8,则x²+y²多少
(x²+y²+1)(x²+y²+3)=8,则x²+y²多少
(x²+y²+1)(x²+y²+3)=8,则x²+y²多少
设X^2+Y2=Z,
于是原方程化为:(Z+1)(Z+3)=8,
Z^2+4Z-5=0,
(Z+5)(Z-1)=0
Z=1或-5,
但X^2+Y^2≠-5,
∴X^2+Y^2=1.