Đáp án:

\(\eqalign{ & a)\,\,P = {{\sqrt {xy} } \over {x - \sqrt {xy} + \sqrt y }} \cr & b)\,\,{{\sqrt 2 } \over 2} \cr & c)\,\,\left| P \right| = P \cr & d)\,\,x = y \cr} \)

Giải thích các bước giải:

$$\eqalign{ & a)\,\,P = \left( {{{x - y} \over {\sqrt x - \sqrt y }} + {{\sqrt {{x^3}} - \sqrt {{y^3}} } \over {y - x}}} \right):\left( {{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + \sqrt {xy} } \over {\sqrt x + \sqrt y }}} \right) \cr & P = \left( {\sqrt x + \sqrt y - {{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)} \over {\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right):\left( {{{x - 2\sqrt {xy} + y + \sqrt {xy} } \over {\sqrt x + \sqrt y }}} \right) \cr & P = \left( {\sqrt x + \sqrt y - {{x + \sqrt {xy} + y} \over {\sqrt x + \sqrt y }}} \right):\left( {{{x - \sqrt {xy} + \sqrt y } \over {\sqrt x + \sqrt y }}} \right) \cr & P = {{{{\left( {\sqrt x + \sqrt y } \right)}^2} - \left( {x + \sqrt {xy} + y} \right)} \over {\sqrt x + \sqrt y }}.{{\sqrt x + \sqrt y } \over {x - \sqrt {xy} + \sqrt y }} \cr & P = {{x + 2\sqrt {xy} + y - x - \sqrt {xy} - y} \over {x - \sqrt {xy} + \sqrt y }} \cr & P = {{\sqrt {xy} } \over {x - \sqrt {xy} + \sqrt y }} \cr & b)\,\,Thay\,\,x = 1;\,\,y = {1 \over 2} \cr & \Rightarrow P = {{\sqrt {{1 \over 2}} } \over {1 - \sqrt {{1 \over 2}} + \sqrt {{1 \over 2}} }} = \sqrt {{1 \over 2}} = {{\sqrt 2 } \over 2} \cr & c)\,\,P = {{\sqrt {xy} } \over {x - \sqrt {xy} + \sqrt y }} \cr & \left\{ \matrix{ \sqrt {xy} \ge 0 \hfill \cr x - \sqrt {xy} + \sqrt y > 0 \hfill \cr} \right.\,\,\forall x;y \ge 0,\,\,x \ne y \cr & \Rightarrow P \ge 0\,\,\forall x;y \ge 0 \Rightarrow \left| P \right| = P \cr & d)\,\,P = {{\sqrt {xy} } \over {x - \sqrt {xy} + \sqrt y }} \cr & P = {1 \over {{{\sqrt x } \over {\sqrt y }} - 1 + {{\sqrt y } \over {\sqrt x }}}} \cr & P \in Z \Rightarrow {{\sqrt x } \over {\sqrt y }} - 1 + {{\sqrt y } \over {\sqrt x }} \in U\left( 1 \right) \Rightarrow \left[ \matrix{ {{\sqrt x } \over {\sqrt y }} - 1 + {{\sqrt y } \over {\sqrt x }} = 1 \hfill \cr {{\sqrt x } \over {\sqrt y }} - 1 + {{\sqrt y } \over {\sqrt x }} = - 1 \hfill \cr} \right. \cr & Ta\,\,co\,\,{{\sqrt x } \over {\sqrt y }} + {{\sqrt y } \over {\sqrt x }} \ge 2\,\,\left( {BDT\,\,Co - si} \right) \cr & \Rightarrow {{\sqrt x } \over {\sqrt y }} - 1 + {{\sqrt y } \over {\sqrt x }} \ge 1 \cr & Dau\,\, = \,\,xay\,\,ra\,\, \Leftrightarrow x = y \cr} $$