a) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{3x}{x-4}\left(dkxd:x\ge0;x\ne4\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{3x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-3x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=-\dfrac{1}{\sqrt{x}+2}\)
Vậy \(P=-\dfrac{1}{\sqrt{x}+2}\).
\(---\)
b) Ta có: \(P\le-\dfrac{1}{3}\Leftrightarrow-\dfrac{1}{\sqrt{x}+2}\le-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}\ge\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}+2\le3\)
\(\Leftrightarrow\sqrt{x}\le1\)
\(\Leftrightarrow x\le1\)
Kết hợp với điều kiện xác định của $x$, ta được:
\(0\le x\le1\)
Vậy với \(0\le x\le1\) thì \(P\le-\dfrac{1}{3}\).
#\(Toru\)
