\[\begin{array}{l} Z = \left( {\frac{{x + 2}}{{\sqrt x + 1}} - \sqrt x } \right):\left( {\frac{{\sqrt x - 4}}{{1 - x}} - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\,\,\,\left( {DK:\,\,x \ge 0;\,\,x \ne 1} \right)\\ = \frac{{x + 2 - x - \sqrt x }}{{\sqrt x + 1}}:\frac{{\sqrt x - 4 - \sqrt x \left( {1 - \sqrt x } \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\ = \frac{{2 - \sqrt x }}{{\sqrt x + 1}}:\frac{{\sqrt x - 4 - \sqrt x + x}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\ = \frac{{2 - \sqrt x }}{{\sqrt x + 1}}:\frac{{x - 4}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\ = \frac{{2 - \sqrt x }}{{\sqrt x + 1}}.\frac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\ = \frac{{1 - \sqrt x }}{{ - \left( {\sqrt x + 2} \right)}} = \frac{{\sqrt x - 1}}{{\sqrt x + 2}}. \end{array}\]

$$\eqalign{ & \left( {{{x + 2} \over {\sqrt x + 1}} - \sqrt x } \right):\left( {{{\sqrt x - 4} \over {1 - x}} - {{\sqrt x } \over {\sqrt x + 1}}} \right) \cr & = {{x + 2 - x - \sqrt x } \over {\sqrt x + 1}}:{{\sqrt x - 4 - \sqrt x \left( {1 - \sqrt x } \right)} \over {\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} \cr & = {{2 - \sqrt x } \over {\sqrt x + 1}}:{{\sqrt x - 4 - \sqrt x + x} \over {\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} \cr & = {{2 - \sqrt x } \over {\sqrt x + 1}}.{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)} \over {x - 4}} \cr & = {{2 - \sqrt x } \over {\sqrt x + 1}}.{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} \cr & = {{\sqrt x - 1} \over {\sqrt x + 2}} \cr} $$