a)
\(ĐKXĐ:\left[{}\begin{matrix}x+3\ne0\\x-3\ne0\\9-x^2\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne-3\\x\ne3\\\left(3-x\right)\left(3+x\right)\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)
b)
\(\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)
\(=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{4}{x-3}\)
b)
thay A=4 ta có
\(4=\dfrac{4}{x-3}\\ =>4\cdot\left(x-3\right)=4\\ =>x-3=1\\ =>x=4\left(tm\right)\)
giúp mk vs ạ ai nhanh mk tick nha 
