a/ ĐK: $x\ge 0,x\ne 1$
$P\,=\dfrac{1}{\sqrt x-1}+\dfrac{1}{\sqrt x+1}+\dfrac{2x}{x-1}\\\quad =\dfrac{\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}+\dfrac{\sqrt x-1}{(\sqrt x-1)(\sqrt x+1)}+\dfrac{2x}{(\sqrt x-1)(\sqrt x+1)}\\\quad =\dfrac{(\sqrt x+1)+(\sqrt x-1)+2x}{(\sqrt x-1)(\sqrt x+1)}\\\quad =\dfrac{\sqrt x+1+\sqrt x -1+2x}{(\sqrt x-1)(\sqrt x+1)}\\\quad =\dfrac{2x+2\sqrt x}{(\sqrt x-1)(\sqrt x+1)}\\\quad =\dfrac{2\sqrt x(\sqrt x+1)}{(\sqrt x-1)(\sqrt x+1)}\\\quad =\dfrac{2\sqrt x}{\sqrt x-1}$
Vậy $P=\dfrac{2\sqrt x}{\sqrt x-1}$
b/ $P>-\dfrac{1}{2}\\↔\dfrac{2\sqrt x}{\sqrt x-1}+\dfrac{1}{2}>0\\↔\dfrac{4\sqrt x}{2(\sqrt x-1)}+\dfrac{\sqrt x-1}{2(\sqrt x-1)}>0\\↔\dfrac{4\sqrt x+\sqrt x-1}{2(\sqrt x-1)}>0\\↔\dfrac{5\sqrt x-1}{\sqrt x-1}>0\\↔\left[\begin{array}{1}\begin{cases}5\sqrt x-1>0\\\sqrt x-1>0\end{cases}\\\begin{cases}5\sqrt x-1<0\\\sqrt x-1<0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}5\sqrt x>1\\\sqrt x>1\end{cases}\\\begin{cases}5\sqrt x<1\\\sqrt x<1\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}\sqrt x>\dfrac{1}{5}\\x>1\end{cases}\\\begin{cases}\sqrt x<\dfrac{1}{5}\\x<1\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}x>\dfrac{1}{25}\\x>1\end{cases}\\\begin{cases}x<\dfrac{1}{25}\\x<1\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}x>1\\x<\dfrac{1}{25}\end{array}\right.$
Kết hợp điều kiện: $x\ge 0,x\ne 1$
$→\left[\begin{array}{1}x>1\\0\le x<\dfrac{1}{25}(x\ne 1)\end{array}\right.$
Vậy $x>1$ hoặc $0\le x<\dfrac{1}{25}$ thì $P>-\dfrac{1}{2}$
