\(\dfrac{1}{3-x}+\dfrac{x}{3+x}=\dfrac{x^2+9}{x^2-9}\)
\(\Leftrightarrow\dfrac{x}{3+x}-\dfrac{1}{x-3}=\dfrac{x^2+9}{\left(x-3\right)\left(x+3\right)}\left(1\right)\)
Đk : (x - 3)(x + 3) \(\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
(1) \(\Leftrightarrow\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow x^2-3x-x-3=x^2+9\)
\(\Leftrightarrow x^2-x^2-3x-x=9+3\)
\(\Leftrightarrow-4x=12\Leftrightarrow x=-3\left(KTMĐK\right)\)
Vậy phương trình trên vô nghiệm \(S=\varnothing\)
