已知(x²+y²)(x²+y²-1)-6=0,则x²+y²的值是
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已知(x²+y²)(x²+y²-1)-6=0,则x²+y²的值是
已知(x²+y²)(x²+y²-1)-6=0,则x²+y²的值是
已知(x²+y²)(x²+y²-1)-6=0,则x²+y²的值是
设t=x²+y²,t≥0
则(x²+y²)(x²+y²-1)-6=0可化为:
t(t-1)-6=0
t²-t-6=0
(t-3)(t+2)=0
则t=3或-2(舍去)
∴x²+y²=3
【中学生数理化】团队wdxf4444为您解答!
将x²+y²看做一个整体
设:x²+y²=a
则:a(a-1)-6=0
化简并整理得:a²-a-6=0
解得:a=3或a=-2
∵x²+y²>0
∴a=-2舍去
∴x²+y²=3
不懂追问~
希望我的回答对你有帮助,采纳吧O(∩_∩)O!
答:
(x²+y²)(x²+y²-1)-6=0
(x²+y²)²-(x²+y²)-6=0
(x²+y²-3)(x²+y²+2)=0
在实数范围内:x²+y²=3
在复数范围内:x²+y²=3或者x²+y²=-2
你好,将所求部分看成一个整体,设为a,则a(a-1)-6=0解得a=2.或-2(舍去)希望可以帮到你,望采纳,谢谢~