a
(a) \(y = {1 \over x} = {1 \over {2 + \sqrt 3 }} = {{2 - \sqrt 3 } \over {4 - 3}} = 2 - \sqrt 3 \)

\(\therefore {x \over {\sqrt 2 + \sqrt x }} + {y \over {\sqrt 2 - \sqrt y }}\)=\({{x(\sqrt 2 - \sqrt x )} \over {2 - x}} + {{y(\sqrt 2 + \sqrt y )} \over {2 - y}}\)

==> \({{x(\sqrt x - \sqrt 2 )} \over {x - 2}} + {{y\,(\sqrt y + \sqrt 2 )} \over {2 - y}} = {{x\,(\sqrt x - \sqrt 2 )} \over {\sqrt 3 }} + {{y\,(\sqrt y + \sqrt 2 )} \over {\sqrt 3 }}\)

= \({1 \over {\sqrt 3 }}\,[x\sqrt x + y\sqrt y + \sqrt 2 (y - x)]\)

= \({1 \over {\sqrt 3 }}\,[{(2 + \sqrt 3 )^{3/2}} + {(2 - \sqrt 3 )^{3/2}} + \sqrt 2 ( - 2\sqrt 3 )]\)

= \({1 \over {\sqrt 3 }}\left[ {{1 \over {{2^{3/2}}}}{{(4 + 2\sqrt 3 )}^{3/2}} + {1 \over {{2^{3/2}}}}{{(4 - 2\sqrt 3 )}^{3/2}} - 2\sqrt 6 } \right]\)

= \({1 \over {\sqrt 3 }}\,\left[ {{1 \over {2\sqrt 2 }}\{ {{(\sqrt 3 + 1]}^3} + {{(\sqrt 3 - 1)}^3}\} - 2\sqrt 6 } \right]\)

= \({1 \over {\sqrt 3 }}\left[ {{1 \over {2\sqrt 2 }}\{ 2.3\sqrt 3 + 6.\sqrt 3 \} - 2\sqrt 6 } \right]\)

= \({1 \over {\sqrt 3 }}(3\sqrt 6 - 2\sqrt 6 ) = \sqrt 2 \).