Question

先化简分式(a^2-9)/(a^2+6a+9)+（a-3）/(a^2+3a)-(a-a^2)/(a-1)然后在0,1,2

Answer 1

(a^2-9)/(a^2+6a+9)+（a-3）/(a^2+3a)-(a-a^2)/(a-1)

=(a+3)(a-3)/(a+3)²+(a-3)/[a(a+3)]+a(a-1)/(a-1)

=(a-3)/(a+3)+(a-3)/[a(a+3)]+a

=(a-3)[1/(a+3)+/[a(a+3)]]+a

=(a-3)[(a+1)/[a(a+3)]]+a

当a=2时

原式=(2-3)[(2+1)/[2(2+3)]+2

=-3/10+2

=17/10