$a)B=\dfrac{x+2\sqrt{x}+1}{x\sqrt{x}+1}\\ ĐKXĐ:x\ge 0\\ B=\dfrac{x+2\sqrt{x}+1}{x\sqrt{x}+1}\\ =\dfrac{(\sqrt{x}+1)^2}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\\ =\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\\ x=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)}=(\sqrt{2}-1)^2\\ \Rightarrow B((\sqrt{2}-1)^2)=\dfrac{\sqrt{(\sqrt{2}-1)^2}+1}{(\sqrt{2}-1)^2-\sqrt{(\sqrt{2}-1)^2}+1}\\ =\dfrac{\sqrt{2}-1+1}{3-2\sqrt{2}-\sqrt{2}+1+1}\\ =\dfrac{\sqrt{2}}{5-3\sqrt{2}}\\ =\dfrac{\sqrt{2}(5+3\sqrt{2})}{(5-3\sqrt{2})(5+3\sqrt{2})}\\ =\dfrac{6+5\sqrt{2}}{7}\\ b)P=\dfrac{A}{B}=\dfrac{\dfrac{1-\sqrt{x}}{x-\sqrt{x}+1}}{\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}}\\ =\dfrac{1-\sqrt{x}}{1+\sqrt{x}}\\ c)P>\dfrac{1}{2}\\ \Leftrightarrow \dfrac{1-\sqrt{x}}{1+\sqrt{x}}>\dfrac{1}{2}\\ \Leftrightarrow \dfrac{1-\sqrt{x}}{1+\sqrt{x}}-\dfrac{1}{2}>0\\ \Leftrightarrow \dfrac{2-2\sqrt{x}}{2(1+\sqrt{x})}-\dfrac{1+\sqrt{x}}{2(1+\sqrt{x})}>0\\ \Leftrightarrow \dfrac{2-2\sqrt{x}-1-\sqrt{x}}{2(1+\sqrt{x})}>0\\ \Leftrightarrow \dfrac{1-3\sqrt{x}}{2(1+\sqrt{x})}>0$
$\Leftrightarrow 1-3\sqrt{x}>0($Do $\sqrt{x}+1 \ge 0 \, \forall \, x)$
$\Leftrightarrow x<\dfrac{1}{9}$
Kết hợp ĐK $\Rightarrow 0<x<\dfrac{1}{9}$
