\(a,P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{3x}{x-4}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{3x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-3x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-3x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x-2}\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-1}{\sqrt{x}+2}\)
`b,` Để `P <= -1/3` Thì :
\(-\dfrac{1}{\sqrt{x}+2}\le-\dfrac{1}{3}\\ \Leftrightarrow-\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{3}\le0\\ \Leftrightarrow\dfrac{-3+\sqrt{x}+2}{3\left(\sqrt{x}+2\right)}\le0\\ \Leftrightarrow\dfrac{\sqrt{x}-1}{3\left(\sqrt{x}+2\right)}\le0\\ \Leftrightarrow\sqrt{x}-1\le0\\ \Leftrightarrow\sqrt{x}\le1\\ \Leftrightarrow x\le1\)
Kết hợp với điều kiện ta có :
`0<= x <=1`
