我认为应该是这样解的∵a,b,c 成等比数列
∴b^2=ac
∴(2-logb a)*(1+logc a)
=2+2(logc a)-logb a-(logb a)*(logc a)
=2+2/(loga c) -1/(loga b) - 1/(logb a)*(logc a)
=2+(2loga b - loga c - 1)/(loga b * loga c)
=2+(loga b^2/c -1)/(loga b * loga c)
=2+(loga a -1)/(loga b * loga c)
=2+(1-1)/(loga b * loga c)
=2