$\sqrt {(3 + \sqrt 5 )} $ is equal to
$\sqrt 5 + 1$
$\sqrt 3 + \sqrt 2 $
$(\sqrt 5 + 1)/\sqrt 2 $
${1 \over 2}(\sqrt 5 + 1)$
The square root of $\sqrt {(50)} + \sqrt {(48)} $ is
For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$
If $x + \sqrt {({x^2} + 1)} = a,$ then $x =$
${{{{2.3}^{n + 1}} + {{7.3}^{n - 1}}} \over {{3^{n + 2}} - 2{{(1/3)}^{l - n}}}} = $
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} = $