If ${x^y} = {y^x},$then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ where $k = $

If $x = 3 - \sqrt {5,} $ then ${{\sqrt x } \over {\sqrt 2 + \sqrt {(3x - 2)} }} = $

Solution of the equation  ${4.9^{x - 1}} = 3\sqrt {({2^{2x + 1}})} $ has the solution

${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $

If $x = {{\sqrt 5 + \sqrt 2 } \over {\sqrt 5 - \sqrt 2 }},y = {{\sqrt 5 - \sqrt 2 } \over {\sqrt 5 + \sqrt 2 }},$ then $3{x^2} + 4xy - 3{y^2} = $

${{3\sqrt 2 } \over {\sqrt 6 + \sqrt 3 }} - {{4\sqrt 3 } \over {\sqrt 6 + \sqrt 2 }} + {{\sqrt 6 } \over {\sqrt 3 + \sqrt 2 }} = $