设a/2=b/3=2c/1=k(k≠0)
则a=2k,b=3k,2c=k
2a+3b-2c/2a-b+2c
=(2x2k+3x3k-k)/(2x2k-3k+k)
=(4k+9k-k)/(4k-3k+k)
=12k/2k
=6