已知x：y：z=3:4:5,且x+y+z=12,那么x=?,y=?,z=?

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已知x：y：z=3:4:5,且x+y+z=12,那么x=?,y=?,z=?已知x：y：z=3:4:5,且x+y+z=12,那么x=?,y=?,z=?已知x：y：z=3:4:5,且x+y+z=12,那么

已知x：y：z=3:4:5,且x+y+z=12,那么x=?,y=?,z=?
已知x：y：z=3:4:5,且x+y+z=12,那么x=?,y=?,z=?

已知x：y：z=3:4:5,且x+y+z=12,那么x=?,y=?,z=?
令x：y：z=3:4:5=k
x=3k
y=4k
z=5k
x+y+z=12
3k+4k+5k=12
12k=12
k=1
x=3k=3*1=3
y=4k=4*1=4
z=5k=5*1=5

12/(3+4+5)=1
x=3
y=4
Z=5

x=3 y=4 z=5
解由x：y：z=3:4:5得出 y=4x/3 z=5x/3
带入x+y+z=12 解得x=3
即 y=4x/3 =4 z=5x/3=5

设x=3t,y=4t,z=5t
所以x+y+z=12t=12
所以t=1
所以x=3,y=4,z=5

x=3,y=4,z=5

x：y：z=3:4:5 由此可以设定：
x=3a,
y=4a,
z=5a
由x+y+z=12 推出 3a + 4a + 5a = 12，计算出 a=1.
那么可以得到，x=3,y=4,z=5

设x=3a,y=4a,z=5a
则3a+4a+5a=12
a=1
所以：x=3 y=4 z=5

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