\(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(\)ĐKXĐ : \(\left\{{}\begin{matrix}x-2\sqrt{x}-3\ne0\\3-\sqrt{x}\ne0\end{matrix}\right.\)⇔ \(\left\{{}\begin{matrix}\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\ne0\\\sqrt{x}\ne3\end{matrix}\right.\)⇔ \(x\ne9\)
\(A=\dfrac{x\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x\sqrt{x}+8\sqrt{x}-3x-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}\left(x+8\right)-3\left(x+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x+8}{\sqrt{x}+1}\)
b) \(A=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{4\sqrt{x}+4+x-4\sqrt{x}+4}{\sqrt{x}+1}=4+\dfrac{\left(\sqrt{x}-4\right)^2}{\sqrt{x}+1}\)
Do : \(\dfrac{\left(\sqrt{x}-4\right)^2}{\sqrt{x}+1}\ge0\forall x\)
\(\Rightarrow\) \(\dfrac{\left(\sqrt{x}-4\right)^2}{\sqrt{x}+1}+4\ge4\)
\(\Rightarrow\) \(A_{Min}=4\Leftrightarrow x=16\left(TMĐK\right)\)
