$\begin{array}{l} a)\dfrac{{{x^2} - 3}}{{x + 6\sqrt x + 9}} = \dfrac{{\left( {x - \sqrt 3 } \right)\left( {x + \sqrt 3 } \right)}}{{{{\left( {\sqrt x + 3} \right)}^2}}}\\ \to \text{Không rút gọn được}\\ b)\dfrac{{{x^2} + 2\sqrt 5 x + 5}}{{{x^2} - 5}}\\ = \dfrac{{{{\left( {x + \sqrt 5 } \right)}^2}}}{{\left( {x - \sqrt 5 } \right)\left( {x + \sqrt 5 } \right)}} = \dfrac{{x + \sqrt 5 }}{{x - \sqrt 5 }}\\ 6)\\ a)\dfrac{{\sqrt {3 - 2\sqrt 2 } }}{{\sqrt {17 - 12\sqrt 2 } }} - \dfrac{{\sqrt {3 + 2\sqrt 2 } }}{{\sqrt {17 + 12\sqrt 2 } }}\\ = \dfrac{{\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} }}{{\sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} }} - \dfrac{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} }}{{\sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} }}\\ = \dfrac{{\sqrt 2 - 1}}{{3 - 2\sqrt 2 }} - \dfrac{{\sqrt 2 + 1}}{{3 + 2\sqrt 2 }}\\ = \dfrac{{\sqrt 2 - 1}}{{{{\left( {\sqrt 2 - 1} \right)}^2}}} - \dfrac{{\sqrt 2 + 1}}{{{{\left( {\sqrt 2 + 1} \right)}^2}}}\\ = \dfrac{1}{{\sqrt 2 - 1}} - \dfrac{1}{{\sqrt 2 + 1}}\\ = \sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right)\\ = 2 \end{array}$
