1) điều kiện : \(x>0\)
P = \(\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\) = \(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)
P = \(\dfrac{x-2-\left(\sqrt{x}+2\right)+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\) = \(\dfrac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)}\) = \(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\) = \(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
2) ta có P \(-1\) \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}}-1\) \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2-\sqrt{x}}{\sqrt{x}}\)
\(\Leftrightarrow\) \(\dfrac{-2}{\sqrt{x}}\) ta có \(\sqrt{x}\) \(>0\) (đk : \(x>0\) ) mà \(-2< 0\)
\(\Rightarrow\) \(\dfrac{-2}{\sqrt{x}}< 0\) \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}}-1< 0\) \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}}< 1\) \(\Leftrightarrow\) \(p< 1\)
3) ta có \(P=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{3}\) \(\Leftrightarrow\) \(\sqrt{x}=3\left(\sqrt{x}-2\right)\)
\(\Leftrightarrow\) \(\sqrt{x}=3\sqrt{x}-6\) \(\Leftrightarrow\) \(-2\sqrt{x}=-6\) \(\Leftrightarrow\) \(\sqrt{x}=3\) \(\Leftrightarrow\) \(x=9\) (tmđk)
