Question

已知2/x=3/y=7/z,求xy+xz+yz/x^2+y^2+z^2

Answer 1

令2/x=3/y=7/z=k

∴x=2/k

y=3/k

z=7/k

∴(xy+xz+yz)/(x^2+y^2+z^2)

=(2/k*3/k+2/k*7/k+3/k*7/k)/(4/k²+9/k²+49/k²)

=(6/k²+14/k²+21/k²)/(62/k²)

=(41/k²)/(62/k²)

=41/62