∵(1/a)+(9/b)=1∴9/b=1-(1/a)=(a-1)/a∴b/9=a/(a-1)∴b=(9a)/(a-1)=[9(a-1)+9]/(a-1)=9+[9/(a-1)]∴(a-1)(b-9)=9又a,b∈N∴可得a-1=1且b-9=9.或a-1=9,且b-9=1.或a-1=b-9=3∴(a,b)=(2,18)=(10,10)=(4,12)显然,(a+b)...