ĐKXĐ: \(x\ge0;x\ne1;4\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)}:\left(\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P< 0\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\Rightarrow\sqrt{x}-1< 0\Rightarrow x< 1\Rightarrow0\le x< 1\)
\(P=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)
Do \(\sqrt{x}+2\ge2\Rightarrow\frac{3}{\sqrt{x}+2}\le\frac{3}{2}\Rightarrow P\ge1-\frac{3}{2}=-\frac{1}{2}\)
\(\Rightarrow P_{min}=-\frac{1}{2}\) khi \(x=0\)
