Question

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.［,］

Answer 1

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.