Bài 3:
1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(P=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)
\(=\dfrac{x-4\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
2: Để \(P\ge2\) thì \(\dfrac{1-2\sqrt{x}+4}{\sqrt{x}-2}\ge0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-5}{\sqrt{x}-2}\le0\)
\(\Leftrightarrow4< x\le\dfrac{25}{4}\)