a)
\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\\ =\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\\ =\left(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\\ =\frac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}< \frac{1}{2}\\ \Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac{1}{2}< 0\\ \Leftrightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}< 0\\ \Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+2\right)}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow\sqrt{x}< 9\)
Vậy với \(0\le x< 9;x\ne1;x\ne4\)thì P<\(\frac{1}{2}\)
c)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)
Để P đạt GTNN thì \(\frac{3}{\sqrt{x}+2}\)đạt GTLN \(\Leftrightarrow\sqrt{x}+2\)đạt GTNN
\(\sqrt{x}+2\ge2\forall x\Leftrightarrow\)GTNN là 2 khi x=0
Khi đó, min P = \(1-\frac{3}{2}=-\frac{1}{2}\)
