已知X/3=Y/4=Z/6,求(XY+YZ+XZ)/(X^2+Y^2+Z^2)的值

问题描述:

已知X/3=Y/4=Z/6,求(XY+YZ+XZ)/(X^2+Y^2+Z^2)的值

X/3=Y/4=Z/6=a
XY+YZ+XZ=12a^2+24a^2+18a^2=54a^2
X^2+Y^2+Z^2=9a^2+16a^2=36a^2=61a^2
所以(XY+YZ+XZ)/(X^2+Y^2+Z^2)=54/61
(注:X/3=Y/4=Z/6!=0)

x=3y/4,z=3y/2
原式=(3y&sup2/4+3y&sup2/2+9y&sup2/8)/(9y&sup2/16+y&sup2+9y&sup2/4)
=(3/4+3/2+9/8)/(9/16+1+9/4)
=(27/8)/(61/16)
=54/61