题目内容
若实数x,y,z满足x2+y2+z2=1,则xy+yz+zx的取值范围是A.[-1,1] B.[,1] C.[-1,] D.[,]
答案:B xy+yz+zx≤x2+y2+z2=1.
又2(xy+yz+zx)=(x+y+z)2-(x2+y2+z2)=(x+y+z)2-1≥-1,∴≤xy+yz+zx≤1.
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题目内容
若实数x,y,z满足x2+y2+z2=1,则xy+yz+zx的取值范围是A.[-1,1] B.[,1] C.[-1,] D.[,]
答案:B xy+yz+zx≤x2+y2+z2=1.
又2(xy+yz+zx)=(x+y+z)2-(x2+y2+z2)=(x+y+z)2-1≥-1,∴≤xy+yz+zx≤1.