$\begin{array}{l}A = \left(1 + \dfrac{\sqrt{a}}{a + 1}\right): \left(\dfrac{1}{\sqrt{a} - 1}-\dfrac{2\sqrt{a}}{a\sqrt{a} + \sqrt{a} - a -1}\right) \,\,\\ ĐKXĐ: a \geq 0;\, a \ne 1\\ A = \left(\dfrac{a + 1 + \sqrt{a}}{a + 1}\right):\left(\dfrac{1}{\sqrt{a} - 1} - \dfrac{2\sqrt{a}}{(\sqrt{a}-1)(a + 1)}\right)\\ =\left(\dfrac{a + 1 + \sqrt{a}}{a + 1}\right):\left[\dfrac{a + 1 -2\sqrt{a}}{(\sqrt{a}-1)(a + 1)}\right]\\ = \left(\dfrac{a + 1 + \sqrt{a}}{a + 1}\right):\dfrac{(\sqrt{a} - 1)^2}{(\sqrt{a}-1)(a + 1)}\\ = \left(\dfrac{a + 1 + \sqrt{a}}{a + 1}\right):\dfrac{\sqrt{a} - 1}{a + 1}\\ = \left(\dfrac{a + 1 + \sqrt{a}}{a + 1}\right)\cdot \dfrac{a+1}{\sqrt{a} - 1}\\ = \dfrac{a + 1 + \sqrt{a}}{\sqrt{a} - 1}\\ \\A > 1\\ \Leftrightarrow \dfrac{a + 1 + \sqrt{a}}{\sqrt{a} - 1} > 1\\ \Leftrightarrow \dfrac{a+2}{\sqrt{a}-1} > 0\\ \Leftrightarrow \sqrt{a} > 1\\\Leftrightarrow a > 1\\ \\\text{Ta có: }a = 1995 - 2\sqrt{1994}\\ = 1994 - 2\sqrt{1994} + 1\\ = (\sqrt{1994} - 1)^2\\ \Rightarrow \sqrt{a} = \sqrt{1994} - 1\\ \text{Thay vào A, ta được:}\\ A = \dfrac{1995 - 2\sqrt{1994} + 1 + \sqrt{1994} - 1}{\sqrt{1994} - 1 - 1}\\ = \dfrac{1995 - \sqrt{1994}}{\sqrt{1994}-2} = \dfrac{1995\sqrt{1994} - \sqrt{1994} + 1}{1990} \end{array}$
