a) Ta có: \(A=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\)
b) Để A nguyên thì \(\sqrt{x}+3⋮\sqrt{x}-1\)
\(\Leftrightarrow4⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;3;5\right\}\)
hay \(x\in\left\{0;4;9;25\right\}\)
