Chứng minh đẳng thức:

a). \[\frac{{\sqrt a }}{{\sqrt a  - \sqrt b }} - \frac{{\sqrt b }}{{\sqrt a  + \sqrt b }} - \frac{{2b}}{{a - b}} = 1{\rm{ (a}} \ge {\rm{0}}{\rm{,b}} \ge {\rm{0}}{\rm{,a}} \ne {\rm{0);}}\]

b). \[\frac{{a\sqrt b  + b}}{{a - b}}\sqrt {\frac{{ab + {b^2} - 2\sqrt {a{b^2}} }}{{a\left( {a + 2\sqrt b } \right) + b}}} \left( {\sqrt a  + \sqrt b } \right) = b{\rm{  }}\left( {a > b > 0} \right)\]

a). Ta có \[\frac{{\sqrt a }}{{\sqrt a  - \sqrt b }} - \frac{{\sqrt b }}{{\sqrt a  + \sqrt b }} - \frac{{2b}}{{a - b}}{\rm{   (a}} \ge {\rm{0}}{\rm{,b}} \ge {\rm{0}}{\rm{,a}} \ne {\rm{0);}}\]

\[\begin{array}{l}{\rm{ = }}\frac{{\sqrt a \left( {\sqrt a  + \sqrt b } \right) - \sqrt b \left( {\sqrt a  - \sqrt b } \right)}}{{\left( {\sqrt a  - \sqrt b } \right).\left( {\sqrt a  + \sqrt b } \right)}} - \frac{{2b}}{{a - b}}\\ = \frac{{a + \sqrt {ab}  - \sqrt {ab}  + b}}{{a - b}} - \frac{{2b}}{{a - b}}\end{array}\]

\[ = \frac{{a - b}}{{a - b}} = 1\]

b). Ta có \[\frac{{a\sqrt b  + b}}{{a - b}}\sqrt {\frac{{ab + {b^2} - 2\sqrt {a{b^3}} }}{{a\left( {a + 2\sqrt b } \right) + b}}} \left( {\sqrt a  + \sqrt b } \right){\rm{  }}\left( {a > b > 0} \right)\]

\[\begin{array}{l} = \frac{{\sqrt b \left( {a + \sqrt b } \right)}}{{\left( {\sqrt a  + \sqrt b } \right).\left( {\sqrt a  - \sqrt b } \right)}}\sqrt {\frac{{b\left( {a + b - 2\sqrt {ab} } \right)}}{{{a^2} + 2a\sqrt b  + b}}} \left( {\sqrt a  + \sqrt b } \right)\\ = \frac{{\sqrt b \left( {a + \sqrt b } \right)}}{{\left( {\sqrt a  - \sqrt b } \right)}}.\sqrt {\frac{{b{{\left( {\sqrt a  - \sqrt b } \right)}^2}}}{{{{\left( {a + \sqrt b } \right)}^2}}}} \\ = \frac{{\sqrt b \left( {a + \sqrt b } \right)}}{{\left( {\sqrt a  - \sqrt b } \right)}}.\frac{{\sqrt b \left( {a + \sqrt b } \right)}}{{\left( {\sqrt a  - \sqrt b } \right)}} = b\end{array}\]