Question

已知2分之a=3分之b=4分之c求(a-b-c分之a*2-ab+ac)÷(2分之a-b+c)的值

Answer 1

解：
令a/2=3/b=c/4=t，则a=2t b=3t c=4t
[(a²-ab+ac)/(a-b-c)]÷[(a-b+c)/2]
=[(4t²-6t²+8t²)/(2t-3t-4t)]/[(2t-3t+4t)/2]
=[6t²/(-5t)]/(3t/2)
=(-6t/5)/(3t/2)
=(-6/5)(2/3)
=-4/5

Answer 2

原式=a-b-c分之a(a-b+c)xa-b+c分之2
=2a/a-b-c
设a/2=b/3=c/4=k，
则a=2k,b=3k,c=4k
∴原式=4k/（2k-3k-4k)=-5/4

Answer 3

a\2=b\3=c\4 则a=2b\3 c=4b\3
又因为：(a*2-ab+ac\a-b-c)x2\a-b+c
=2a\a-b-c
=-4\5