初二數(shù)學(xué)競(jìng)賽題2，初二數(shù)學(xué)競(jìng)賽試題1，初二數(shù)學(xué)競(jìng)賽題c/a=(4d-7)/b=(7d-7)/(b+1)由等比定理c/a=(4d-7)/b=(7d-7)/(b+1)=((7d-7)-(4d-7))/((b+1)-1)=3d那么(4d-7)/b=3db=1+(d-7)/3d因?yàn)閎為整數(shù)，所以(d-7)/3d也為整數(shù)設(shè)x=(d-7)/3d,則d=7/(1-3x)因?yàn)閤為整數(shù)而d為正整數(shù)，所以x只能為0或-2,d只能為7或1則b為1或-1但因?yàn)閎為正整數(shù)，所以d=7,b=1,代入c/a=3d則c/a=21,d...