已知x-y=y-z=3/5,x²+y²+z²=1,则xy+yz+xz=?
问题描述：
已知x-y=y-z=3/5,x²+y²+z²=1,则xy+yz+xz=?
答
x-y=3/5
y-z=3/5,
两式相加得
x-z=6/5
(x-y)²+(y-z)²+(x-z)²
=2(x²+y²+z²)-2(xy+yz+xz)
=2-2(xy+yz+xz)
又(x-y)²+(y-z)²+(x-z)²=（3/5）²+（3/5）²+（6/5）²=54/25
2-2(xy+yz+xz)=54/25
xy+yz+xz=-2/25