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)
\(\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)
\(=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{4}{x-3}\)
c) thay P=4 ta có
\(4=\dfrac{4}{x-3}\) \(ĐKXĐ:x\ne3;x\ne-3\)
=> \(4\cdot\left(x-3\right)=4\\ x-3=1\\ x=4\left(tm\right)\)
