a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
b) Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{3x-8\sqrt{x}+27}{9-x}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{3x-8\sqrt{x}+27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+5\sqrt{x}+6+2x-6\sqrt{x}-3x+8\sqrt{x}-27}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{7\sqrt{x}-21}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{7\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{7}{\sqrt{x}+3}\)
c) Để P nhận giá trị nguyên thì \(7⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3\inƯ\left(7\right)\)
\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
mà \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}+3=7\)
\(\Leftrightarrow\sqrt{x}=4\)
hay x=16(nhận)
Vậy: Để P nguyên thì x=16
