已知实数xyz满足x+y+z=0,x2+y2+z2=1,求x的取值范围
1个回答
∵x+y+z=0
∴z=-x-y
∵x2+y2+z2=1
∴x²+y²+x²+2xy+y²=1
2y²+2xy+(2x²-1)=0
∴判别式=4x²-16x²+8≥0
∴-√6/3≤x≤√6/3
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