Phương pháp giải:

Lời giải chi tiết:

$\begin{array}{l} \sqrt {{{\left( {2 - \sqrt 5 } \right)}^2}} - \sqrt {\dfrac{8}{{7 - 3\sqrt 5 }}} \\ = \left| {2 - \sqrt 5 } \right| - \sqrt {\dfrac{{16}}{{14 - 6\sqrt 5 }}} \\ = \left| {2 - \sqrt 5 } \right| - \dfrac{{\sqrt {16} }}{{\sqrt {14 - 6\sqrt 5 } }}\\ = - \left( {2 - \sqrt 5 } \right) - \dfrac{4}{{\sqrt {{3^2} - 2.3.\sqrt 5 + {{\sqrt 5 }^2}} }}\left( {Do{\rm{ }}2 - \sqrt 5 < 0} \right)\\ = - 2 + \sqrt 5 - \dfrac{4}{{\sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} }}\\ = - 2 + \sqrt 5 - \dfrac{4}{{3 - \sqrt 5 }}\\ = - 2 + \sqrt 5 - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}\\ = - 2 + \sqrt 5 - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{4}\\ = - 2 + \sqrt 5 - 3 - \sqrt 5 = - 5 \end{array}$

b)

$\begin{array}{l} A = \dfrac{x}{{\sqrt x - 1}} - \dfrac{{2x - \sqrt x }}{{x - \sqrt x }}\\ = \dfrac{x}{{\sqrt x - 1}} - \dfrac{{\sqrt x \left( {2\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\ = \dfrac{x}{{\sqrt x - 1}} - \dfrac{{2\sqrt x - 1}}{{\sqrt x - 1}}\\ = \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x - 1}}\\ = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}} = \sqrt x - 1 \end{array}$