a. Ta có :
x2-9= (x-3)(x+3)
=> Mtc : (x-3)(x+3)
Nhân tử phụ 1 : (x-3)(x+3):(x-3)(x+3)=1
Nhân tử phụ 2 : (x-3)(x+3):(x+3)=x-3
Qui đồng:
x+15/x2-9= x+15/(x-3)(x+3)=(x-15).1/(x-3)(x+3).1=x-15/(x-3)(x+3)
2/x+3=2.(x-3)/(x+3)(x-3)=2(x-3)/(x-3)(x+3)
x-15/(x-3)(x+3) + 2(x-3)/(x-3)(x+3)
= (x-15)+2(x-3)/(x-3)(x+3)
= x-15+2/x+3
= x-13/x+3