x/3=y/2=z/5,求xy+yz+xz/x²+y²+z²

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x/3=y/2=z/5,求xy+yz+xz/x²+y²+z²
x/3=y/2=z/5,求xy+yz+xz/x²+y²+z²

x/3=y/2=z/5,求xy+yz+xz/x²+y²+z²
令 x/3=y/2=z/5=k
则 x=3k y=2k z=5k
∴(xy+yz+zx)/(x²+y²+z²)
=(6+10+15)k²/(9+4+25)l²
=31/38

你好!
数学之美团为你解答
设 x/3 = y/2 = z/5 = a
则 x=3a,y=2a,z=5a
原式 = (6a²+10a²+15a²) / (9a²+4a²+25a²)
= 31/38

31除以38!!!

设:X/3=Y/2=Z/5=K (X、Y、Z、K不为零)
X=3K Y=2K Z=5K

代入后,原式=(6KK+10KK+15KK)/(9KK+4KK+25KK)=31/38

设X=3A,Y=2A,Z=5A
则(6+10+15)/(9+4+25)=31/38