a) điều kiện xác định : \(x\ge0;x\ne\dfrac{9}{4}\)
ta có : \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow P=\left(\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{2x+3\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}\right)=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)b) thay \(x=\dfrac{3-2\sqrt{2}}{4}=\left(\dfrac{\sqrt{2}-1}{2}\right)^2\) vào \(P\) ta có :
\(P=\dfrac{3\sqrt{\left(\dfrac{\sqrt{2}-1}{2}\right)^2}-5}{2\sqrt{\left(\dfrac{\sqrt{2}-1}{2}\right)^2}+1}=\dfrac{3\left(\dfrac{\sqrt{2}-1}{2}\right)-5}{2\left(\dfrac{\sqrt{2}-1}{2}\right)+1}=\dfrac{6-13\sqrt{2}}{4}\)
c) ta có : \(P-\dfrac{3}{2}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}-\dfrac{3}{2}=\dfrac{3\sqrt{x}-5-3\sqrt{x}-\dfrac{3}{2}}{2\sqrt{x}+1}\)
\(=\dfrac{\dfrac{-13}{2}}{2\sqrt{x}+1}< 0\) \(\Rightarrow P< \dfrac{3}{2}\)
