\(P=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{6-\sqrt{x}}{4-x}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-\sqrt{x}-6+2x-4\sqrt{x}-\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
P<2
=>P-2<0
=>\(\dfrac{3\sqrt{x}}{\sqrt{x}+2}-2< 0\)
=>\(\dfrac{3\sqrt{x}-2\sqrt{x}-4}{\sqrt{x}+2}< 0\)
=>\(\sqrt{x}-4< 0\)
=>\(\sqrt{x}< 4\)
=>0<=x<16
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< 16\\x\ne4\end{matrix}\right.\)
\(P=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{6-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-\sqrt{x}-6+2x-4\sqrt{x}+6-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Để \(P< 2\Rightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}< 2\)
\(\Rightarrow3\sqrt{x}< 2\left(\sqrt{x}+2\right)\) (do \(\sqrt{x}+2>0;\forall x\ge0\))
\(\Leftrightarrow\sqrt{x}< 4\)
\(\Rightarrow x< 16\)
Kết hợp ĐKXĐ \(\Rightarrow\left\{{}\begin{matrix}0\le x< 16\\x\ne4\end{matrix}\right.\)
