已知有理数xy满足(x²+y²)(x²+y²-1)-6=0,则x²+y²的值为
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![已知有理数xy满足(x²+y²)(x²+y²-1)-6=0,则x²+y²的值为](/uploads/image/z/2575572-60-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%9C%89%E7%90%86%E6%95%B0xy%E6%BB%A1%E8%B6%B3%EF%BC%88x%26%23178%3B%2By%26%23178%3B%EF%BC%89%EF%BC%88x%26%23178%3B%2By%26%23178%3B-1%EF%BC%89-6%3D0%2C%E5%88%99x%26%23178%3B%2By%26%23178%3B%E7%9A%84%E5%80%BC%E4%B8%BA)
已知有理数xy满足(x²+y²)(x²+y²-1)-6=0,则x²+y²的值为
已知有理数xy满足(x²+y²)(x²+y²-1)-6=0,则x²+y²的值为
已知有理数xy满足(x²+y²)(x²+y²-1)-6=0,则x²+y²的值为
设t=x²+y² t≥0则
(x²+y²)(x²+y²-1)-6=0
t(t-1)-6=0
t²-t-6=0
(t-3)(t+2)=0
t-3=0
t=3
所以x²+y²的值为3
(x²+y²)(x²+y²-1)-6=0
(x²+y²)²-(x²+y²)-6=0
[(x²+y²)+2][(x²+y²)-3]=0
x²+y²=-2或x²+y²=3
用整体法,把x^2+y^2看做一个整体t,相当于解一元二次方程,那么有t(t-1)-6=0;得t=-2(舍去,因为x^2+y^2>=0),t=3.