ĐKXĐ: \(x>0;x\ne1\)
\(P=\left(1-\sqrt{x}\right)\left(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\left(\frac{x+\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(=\left(1-\sqrt{x}\right)\left[\frac{\sqrt{x}}{\left(1-\sqrt{x}\right)\sqrt{x}}-\frac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\sqrt{x}}\right]\frac{x+\sqrt{x}-2\sqrt{x}+1}{2\sqrt{x}-1}\)
\(=\left(1-\sqrt{x}\right).\frac{2\sqrt{x}-1}{\left(1-\sqrt{x}\right)\sqrt{x}}.\frac{x-\sqrt{x}+1}{2\sqrt{x}-1}\)
\(=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
a, \(x=7-4\sqrt{3}=3-4\sqrt{3}+4=\left(\sqrt{3}-2\right)^2\Rightarrow\sqrt{x}=\sqrt{3}-2\)
\(\Rightarrow P=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\frac{7-4\sqrt{3}-\sqrt{3}+2+1}{\sqrt{3}-2}=\frac{10-5\sqrt{3}}{\sqrt{3}-2}=-2\)
b, \(P>1\Leftrightarrow\frac{x-\sqrt{x}+1}{\sqrt{x}}>1\Leftrightarrow x-\sqrt{x}+1>\sqrt{x}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0,\forall x>0;x\ne1\)
