问题
填空题
设
|
答案
∵
=AB
-OB
=(a-1,1),OA
=AC
-OC
=(-b-1,2).OA
又∵A、B、C三点共线,∴
∥AB
,从而(a-1 )×2-1×(-b-1)=0,AC
∴2a+b=1.
4a+21+b=22a+21+b≥2
=222a+1+b
=44
故4a+21+b的最小值是4,
故答案为:4.
设
|
∵
=AB
-OB
=(a-1,1),OA
=AC
-OC
=(-b-1,2).OA
又∵A、B、C三点共线,∴
∥AB
,从而(a-1 )×2-1×(-b-1)=0,AC
∴2a+b=1.
4a+21+b=22a+21+b≥2
=222a+1+b
=44
故4a+21+b的最小值是4,
故答案为:4.