1: Ta có: \(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-\left(x-9\right)+2x-5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-3\sqrt{x}-7+2x-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-3}\)
2: Để A<1 thì \(\frac{\sqrt{x}-1}{\sqrt{x}-3}< 1\)
\(\Leftrightarrow\sqrt{x}-1< \sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-1-\sqrt{x}+3< 0\)
\(\Leftrightarrow2< 0\)(vô lý)
Vậy: Không có giá trị nào của x thỏa mãn A<1
3: Để A nhận giá trị nguyên thì \(\sqrt{x}-1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3+2⋮\sqrt{x}-3\)
mà \(\sqrt{x}-3⋮\sqrt{x}-3\)
nên \(2⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\inƯ\left(2\right)\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1\right\}\)
\(\Leftrightarrow x\in\left\{16;4;25;1\right\}\)
mà \(x\ge0\) và \(x\notin\left\{4;9\right\}\)(ĐKXĐ)
nên \(x\in\left\{16;25;1\right\}\)
Vậy: để A nhận giá trị nguyên thì \(x\in\left\{16;25;1\right\}\)
