举报
zhenyun1982
抱歉。 a=根号(-2b+14)+根号(b-7)+3 则 b=7 a=3 a=根号(2b-14)+根号(b-7)+3 a=(根号2+1)根号(b-7)+3 b-a=b-(根号2+1)根号(b-7)-3 =(b-7)-(根号2+1)根号(b-7)+4 =根号(b-7)^2-2[(根号2+1)/2]+[(根号2+1)/2]^2+4-[(根号2+1)/2]^2 =[根号(b-7)-(根号2+1)/2]^2+4-(3+2根号2)/4 =[根号(b-7)-(根号2+1)/2]^2+(13-2根号2)/4 所以 b-a>0 根号[(a-b)^2]=根号[(b-a)^2]=b-a=[根号(b-7)-(根号2+1)/2]^2+(13-2根号2)/4