Question

求使b-ab分之a的平方+ab除以2分之a+b乘a的平方-ab分之b的平方-ab具有正整数值的所有a的整数值

Answer 1

(a²+ab)/(b-ab)÷(a+b)/2×(b²-ab)/(a²-ab)

=a(a+b)/b(1-a)×2/(a+b)×[-b(a-b)/a(a-b)]

=-2a/(1-a)

=2a/(a-1)

=2(a-1+1)/(a-1)

=2[(a-1)/(a-1)+1/(a-1)]

=2[1+1/(a-1)]

=2+2/(a-1)是整数

则a-1是2的约数

所以a-1=1,-1,2,-2

a=2,a=0,a=3,a=-1

a²-ab=a(a-b)在分母

所以a≠0

所以a=2,a=3,a=-1