x+y+z=2y=3z(y≠0),求（xy+yz+zx）÷（x²+y²+z²）的值.
x+y+z=2y=3z(y≠0),求（xy+yz+zx）÷（x²+y²+z²）的值.
數(shù)學(xué)人氣：277 ℃時(shí)間：2020-06-17 11:45:28
優(yōu)質(zhì)解答
令x+y+z=2y=3z=6ky = 3kz = 2kx = k（xy+yz+zx）÷（x²+y²+z²）=(3k²+6k²+2k²)÷(k²+9k²+4k²)=(11k²)÷(14k²)=11/14
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