\(\begin{array}{l}
f)\\
\quad F = \left(\dfrac{\sqrt x}{\sqrt x - 1} - \dfrac{1}{x - \sqrt x}\right):\left(\dfrac{1}{\sqrt x + 1} + \dfrac{2}{x - 1}\right)\quad (x >0; x \ne 1)\\
\to F = \left[\dfrac{\sqrt x}{\sqrt x - 1} - \dfrac{1}{\sqrt x\left(\sqrt x - 1\right)}\right]:\left[\dfrac{1}{\sqrt x + 1} + \dfrac{2}{\left(\sqrt x + 1\right)\left(\sqrt x - 1\right)}\right]\\
\to F = \dfrac{x - 1}{\sqrt x\left(\sqrt x - 1\right)}:\dfrac{\sqrt x - 1 + 2}{\left(\sqrt x + 1\right)\left(\sqrt x - 1\right)}\\
\to F = \dfrac{\left(\sqrt x + 1\right)\left(\sqrt x - 1\right)}{\sqrt x\left(\sqrt x - 1\right)}:\dfrac{\sqrt x + 1}{\left(\sqrt x + 1\right)\left(\sqrt x - 1\right)}\\
\to F = \dfrac{\sqrt x + 1}{\sqrt x}:\dfrac{1}{\sqrt x - 1}\\
\to F = \dfrac{\left(\sqrt x + 1\right)\left(\sqrt x- 1\right)}{\sqrt x}\\
\to F = \dfrac{ x-1}{\sqrt x}\\
g)\\
\quad G = \left[\dfrac{x + 3\sqrt x+ 2}{\left(\sqrt x + 2\right)\left(\sqrt x - 1\right)} - \dfrac{x + \sqrt x}{x - 1}\right]:\left(\dfrac{1}{\sqrt x+ 1} + \dfrac{1}{\sqrt x - 1}\right)\quad (x> 0;x\ne 1)\\
\to G = \left[\dfrac{\left(\sqrt x +2\right)\left(\sqrt x + 1\right)}{\left(\sqrt x + 2\right)\left(\sqrt x - 1\right)} - \dfrac{\sqrt x\left(\sqrt x +1\right)}{\left(\sqrt x+ 1\right)\left(\sqrt x - 1\right)}\right]:\left[\dfrac{\sqrt x+ 1 + \sqrt x- 1}{\left(\sqrt x+ 1\right)\left(\sqrt x - 1\right)}\right]\\
\to G = \left(\dfrac{\sqrt x +1}{\sqrt x - 1} - \dfrac{\sqrt x}{\sqrt x-1}\right):\dfrac{2\sqrt x}{\left(\sqrt x+ 1\right)\left(\sqrt x - 1\right)}\\
\to G = \dfrac{1}{\sqrt x- 1}\cdot \dfrac{\left(\sqrt x+ 1\right)\left(\sqrt x - 1\right)}{2\sqrt x}\\
\to G = \dfrac{\sqrt x + 1}{2\sqrt x}
\end{array}\)