3a-6b+6/2cd+3a已知;a与b互为相反数c与d互为倒数且3a+2不等于0

问题描述:

3a-6b+6/2cd+3a已知;a与b互为相反数c与d互为倒数且3a+2不等于0

a与b互为相反数,则a+b=0 b=-a
c与d互为倒数,则cd=1
且3a+2不等于0
所以(3a-6b+6)/(2cd+3a)
=[3a-6(-a)+6]/(3a+2)
=(9a+6)/(3a+2)
=3(3a+2)/(3a+2)
=3