a)Đk \(x\ge0,x\ne1\)
\(\Rightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b)\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)
\(\Rightarrow P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2-\sqrt{3}-1}{2-\sqrt{3}+1}=\dfrac{1-\sqrt{3}}{3-\sqrt{3}}=\dfrac{-\sqrt{3}}{3}\)
c)\(TacoP>\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}>\dfrac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}-2>\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
Vậy\(x>9\)Thì \(P>\dfrac{1}{2}\)
c.
\(P< \frac{1}{2}\Leftrightarrow \frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Leftrightarrow \frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\Leftrightarrow \frac{\sqrt{x}-3}{\sqrt{x}+1}< 0\)
$\Leftrightarrow \sqrt{x}-3< 0\Leftrightarrow 0\leq x< 9$
Kết hợp đkxđ suy ra $0\leq x< 9; x\neq 1$
