a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)
\(A=\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}-\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(=\dfrac{x-2\sqrt{x}-x+4}{x-4}:\left(\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(=\dfrac{-2\sqrt{x}+4}{x-4}:\left(\dfrac{4-x+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}+2}\right)\)
\(=\dfrac{-2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}+2}\)
\(=\dfrac{-2}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+2}{-\left(\sqrt{x}-3\right)}=\dfrac{2}{\sqrt{x}-3}\)
b: Để A là số nguyên thì \(\sqrt{x}-3\inƯ\left(2\right)\)
=>\(\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1\right\}\)
=>\(x\in\left\{16;4;25;1\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{1;16;25\right\}\)
