1: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
\(P=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{9-x}\right):\dfrac{2\sqrt{x}+1}{6-\sqrt{4x}}\)
\(=\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{-2\left(\sqrt{x}-3\right)}{2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-3-x-7\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{-2\left(\sqrt{x}-3\right)}{2\sqrt{x}+1}\)
\(=\dfrac{-x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)}\cdot\dfrac{-2}{2\sqrt{x}+1}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)^2}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}+3\right)}{2\sqrt{x}+1}=\dfrac{2\sqrt{x}+6}{2\sqrt{x}+1}\)
2: Để P là số nguyên thì \(2\sqrt{x}+6⋮2\sqrt{x}+1\)
=>\(2\sqrt{x}+1+5⋮2\sqrt{x}+1\)
=>\(5⋮2\sqrt{x}+1\)
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
=>\(2\sqrt{x}\in\left\{0;4\right\}\)
=>\(\sqrt{x}\in\left\{0;2\right\}\)
=>\(x\in\left\{0;4\right\}\)
Kết hợp ĐKXĐ, ta có: \(x\in\left\{0;4\right\}\)
