$\begin{array}{l} 1)\dfrac{2}{{\sqrt {13} - \sqrt {11} }} + \dfrac{{10}}{{\sqrt {11} - 1}} + \dfrac{9}{{\sqrt {13} - 2}}\\ = \dfrac{{2\left( {\sqrt {13} + \sqrt {11} } \right)}}{{13 - 11}} + \dfrac{{10\left( {\sqrt {11} + 1} \right)}}{{11 - 1}} + \dfrac{{9\left( {\sqrt {13} + 2} \right)}}{{13 - 4}}\\ = \sqrt {13} + \sqrt {11} + \sqrt {11} + 1 + \sqrt {13} + 2\\ = 2\sqrt {13} + 2\sqrt {11} + 3\\ 2)\dfrac{2}{{\sqrt 7 - \sqrt 5 }} + \dfrac{6}{{\sqrt {11} + \sqrt 5 }}\\ = \dfrac{{2\left( {\sqrt 7 + \sqrt 5 } \right)}}{{7 - 5}} + \dfrac{{6\left( {\sqrt {11} - \sqrt 5 } \right)}}{{11 - 5}}\\ = \sqrt 7 + \sqrt 5 + \sqrt {11} - \sqrt 5 = \sqrt {11} + \sqrt 7 \\ 3)\dfrac{6}{{4 + \sqrt {4 - 2\sqrt 3 } }} = \dfrac{4}{{4 + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }} = \dfrac{4}{{4 + \sqrt 3 - 1}} = \dfrac{4}{{3 + \sqrt 3 }}\\ = \dfrac{{4\left( {3 - \sqrt 3 } \right)}}{{9 - 3}} = \dfrac{{2\left( {3 - \sqrt 3 } \right)}}{3}\\ 4)\dfrac{1}{{\sqrt {7 - 2\sqrt 6 } + 1}} - \dfrac{1}{{\sqrt {7 + 2\sqrt 6 } + 1}}\\ = \dfrac{1}{{\sqrt {{{\left( {\sqrt 6 - 1} \right)}^2}} + 1}} - \dfrac{1}{{\sqrt {{{\left( {\sqrt 6 + 1} \right)}^2}} + 1}}\\ = \dfrac{1}{{\sqrt 6 - 1 + 1}} - \dfrac{1}{{\sqrt 6 + 1 + 1}} = \dfrac{1}{{\sqrt 6 }} - \dfrac{1}{{\sqrt 6 + 2}}\\ = \dfrac{{\sqrt 6 }}{6} - \dfrac{{\sqrt 6 - 2}}{{6 - 4}} = \dfrac{{\sqrt 6 }}{6} - \dfrac{{\sqrt 6 - 2}}{2} = \dfrac{{\sqrt 6 - 3\sqrt 6 + 6}}{6}\\ = \dfrac{{6 - 2\sqrt 6 }}{6} = \dfrac{{3 - \sqrt 6 }}{3}\\ 5)\dfrac{{\sqrt 3 }}{{\sqrt {\sqrt 3 + 1} - 1}} - \dfrac{{\sqrt 3 }}{{\sqrt {\sqrt 3 + 1} + 1}}\\ = \dfrac{{\sqrt 3 \left( {\sqrt {\sqrt 3 + 1} + 1} \right) - \sqrt 3 \left( {\sqrt {\sqrt 3 + 1} - 1} \right)}}{{\left( {\sqrt {\sqrt 3 + 1} - 1} \right)\left( {\sqrt {\sqrt 3 + 1} + 1} \right)}}\\ = \dfrac{{\sqrt {3 + \sqrt 3 } + \sqrt 3 - \sqrt {3 + \sqrt 3 } + \sqrt 3 }}{{\sqrt 3 + 1 - 1}} = \dfrac{{2\sqrt 3 }}{{\sqrt 3 }} = 2 \end{array}$
