ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)
a: \(P=\dfrac{x\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{x\sqrt{x}-3-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x+8}{\sqrt{x}+1}\)
b: Khi \(x=14-6\sqrt{5}\) thì \(P=\dfrac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\dfrac{22-6\sqrt{5}}{4-\sqrt{5}}=\dfrac{58-2\sqrt{5}}{11}\)
e: Để P nguyên thì \(x+8⋮\sqrt{x}+1\)
=>\(x-1+9⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;3;9\right\}\)
=>\(\sqrt{x}\in\left\{0;2;8\right\}\)
=>\(x\in\left\{0;4;64\right\}\)
