a) Với \(x\ge0;x\ne1\) ta có:
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Vậy \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\).
b) Thay \(x=9\left(\text{TMĐK}\right)\) vào P, ta được:
\(P=\dfrac{\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
Vậy \(P=\dfrac{1}{2}\) khi \(x=9\).
c) Để \(P< \dfrac{1}{2}\) thì:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\Leftrightarrow\dfrac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}< \dfrac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}\)
\(\Rightarrow2\left(\sqrt{x}-1\right)< \sqrt{x}+1\)
\(\Leftrightarrow2\sqrt{x}-2< \sqrt{x}+1\)
\(\Leftrightarrow2\sqrt{x}-\sqrt{x}< 2+1\)
\(\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)
Vậy để \(P< \dfrac{1}{2}\) thì \(x< 9.\)
