Câu 3
A) ĐKXĐ : (x-3)(x+3) ≠ 0
x+3≠0
x-3≠0
B) MC : (x+3)(X-3)
A= \(\dfrac{3.\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\) + \(\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)+\(\dfrac{18}{x^2-9}\)= \(\dfrac{3x-9+x+3+18}{MC}\)= \(\dfrac{4x+12}{MC}\) =\(\dfrac{\text{4(x+3)}}{MC}\)=\(\dfrac{4}{x-3}\)
c) Tại x=1
Có \(\dfrac{4}{1-3}\)=\(\dfrac{4}{-2}\)=-2