已知实数xyz满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3求x+y+z的值

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已知实数xyz满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3求x+y+z的值

已知实数xyz满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3求x+y+z的值
已知实数xyz满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3求x+y+z的值

已知实数xyz满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3求x+y+z的值
当x=y=z=0时,等式成立,此时x+y+z=0
当x,y,z均不为0时(显然x,y,z不可能部分为0,否则等式一定不可能成立)
由x/(x+1)=y/(y+2)=z/(z+3)知:(x+1)/x=(y+2)/y=(z+3)/z
即1/x=2/y=3/z
令:x=k,y=2k,z=3k
则有(x+y+z)/3=2k
且k/(K+1)=2k
解得:k=1
所以有x+y+z=6k=6

正解!

没说条件啊...x=y=z=0 就满足的啊...

设k=x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3
可得:x=k/(1-k);y=2k/(1-k);z=3k/(1-k)代入k=(x+y+z)/3得6k/2(1-k)=k
解得k=-1
x+y+z=k/(1-k)+2k/(1-k)+3k/(1-k)=6k/(1-k)=-3

设x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3=k
则(x+y+z)k+6k=x+y+z=3k
则(x+y+z)k=-3k
所以x+y+z=-3