已知(x²+y²)(x²-1+y²)=12,求x²+y²的值
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![已知(x²+y²)(x²-1+y²)=12,求x²+y²的值](/uploads/image/z/1002614-14-4.jpg?t=%E5%B7%B2%E7%9F%A5%28x%26%23178%3B%2By%26%23178%3B%29%28x%26%23178%3B-1%2By%26%23178%3B%EF%BC%89%3D12%2C%E6%B1%82x%26%23178%3B%2By%26%23178%3B%E7%9A%84%E5%80%BC)
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已知(x²+y²)(x²-1+y²)=12,求x²+y²的值
已知(x²+y²)(x²-1+y²)=12,求x²+y²的值
已知(x²+y²)(x²-1+y²)=12,求x²+y²的值
(x²+y²)(x²-1+y²)=12
(x²+y²)[(x²+y²)-1]-12=0
(x²+y²)²-(x²+y²)-12=0
[(x²+y²)+3][(x²+y²)-4]=0
x²+y²=-3(舍去)或x²+y²=4
∴x²+y²=4
用换元法计算,设Z=x2+y2,就得出Z的平方-Z-12=0解一元二次方程(Z-4)(Z+3)=0,而x2+y2不可能小于0,所以Z=4即x2+y2=4
希望可以采纳
就是4,把
x²+y²
看做整体就行了