已知x/2=y/3=z/4,求xy+yz+zx/5x的方+3y的方+z的方

网友回答

令x/2=y/3=z/4=k
x=2ky=3kz=4k(xy+yz+zx)/(5x^2+3y^2+z^2)
=(2k*3k+3k*4k+4k*2k)/[5*(2k)^2+3*(3k)^2+(4k^2)]
=(6k^2+12k^2+8k^2)/(20k^2+27k^2+16k^2)
=26k^2/63k^2
=26/63
======以下答案可供参考======
供参考答案1:
设x/2=y/3=z/4=k即可得结果哦26/63