已知x/2=y/3=z/4求(1)x+y+z/x(2)x+y+z/x+y-z(3)y+z-x/z+x-y
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令x/2=y/3=z/4=k
x=2k
y=3k
z=4k
(x+y+z)/x
=(2k+3k+4k)/2k
=9k/2k
=9/2
(x+y+z)/(x+y-z)
=(2k+3k+4k)/(2k+3k-4k)
=9k/k
=9
(y+z-x)/(z+x-y)
=(3k+4k-2k)/(4k+2k-3k)
=5k/3k
=5/3
x=2k
y=3k
z=4k
(x+y+z)/x
=(2k+3k+4k)/2k
=9k/2k
=9/2
(x+y+z)/(x+y-z)
=(2k+3k+4k)/(2k+3k-4k)
=9k/k
=9
(y+z-x)/(z+x-y)
=(3k+4k-2k)/(4k+2k-3k)
=5k/3k
=5/3
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