令a=X/2=Y/3=Z/4X=2aY=3aZ=4a所以原式=(4a+9a-4a)/(2a+6a-12a)=9a/(-4a)=-9/4
X/2=Y/3=Z/4=k(k≠0)x=2k,y=3k,z=4k=(4k+9k-4k)/(2k+6k-12k)=
解：假设X=2 Y=3 Z=4 ( 2x+3y-z)/(x+2y-3z)=(2乘以2+3乘以3-4)除以(2+2乘以3-3乘以4)=-9/4
