ĐKXĐ: \(\left[{}\begin{matrix}x+3\ne0\\x-3\ne0\\9-x^2\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
a, Ta có: \(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\\A=\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{1\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\\ A=\dfrac{3\left(x-3\right)+\left(x+3\right)+18}{\left(x-3\right)\left(x+3\right)}\\ A=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\\ A=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3} \)
b, \(A=4\\ < =>\dfrac{4}{x-3}=4\\ =>4\left(x-3\right)=4\\ < =>4x-12=4\\ < =>4x=16\\ =>x=\dfrac{16}{4}=4\left(TMĐK\right)\)
Vật: Để A=4 thì x=4
