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) \(\left( {\sqrt {2x}  - 1} \right)\left( {2x + \sqrt {2x}  + 1} \right) = \left( {\sqrt {2x}  - 1} \right)\left( {{{\left( {\sqrt {2x} } \right)}^2} + \sqrt {2x} .1 + {1^2}} \right) = {\left( {\sqrt {2x} } \right)^3} - 1 = 2x\sqrt 2 x - 1\).

b) \(\left( {\sqrt x  + 2\sqrt y } \right)\left( {x - 2\sqrt {xy}  + 4y} \right) = \left( {\sqrt x  + 2\sqrt y } \right)\left[ {{{\left( {\sqrt x } \right)}^2} - \sqrt x .2\sqrt y  + {{\left( {2\sqrt y } \right)}^2}} \right]\)

\( = {\left( {\sqrt x } \right)^3} + {\left( {2\sqrt y } \right)^3} = x\sqrt x  + 8y\sqrt y \).

c) \(\left( {\sqrt x  + \sqrt y } \right)\left( {2\sqrt x  - \sqrt y } \right) = \sqrt x .2\sqrt x  - \sqrt x .\sqrt y  + \sqrt y .2\sqrt x  - \sqrt y .\sqrt y \)

\( = 2x - \sqrt {xy}  + 2\sqrt {xy}  - y = 2x + \sqrt {xy}  - y\).

a). Ta có: \(\frac{{\sqrt 2  - 1}}{{\sqrt 2  + 2}} - \frac{2}{{2 + \sqrt 2 }} + \frac{{\sqrt 2  + 1}}{{\sqrt 2 }} = \frac{{\sqrt 2  - 1}}{{\sqrt 2 (1 + \sqrt 2 )}} - \frac{2}{{\sqrt 2 (\sqrt 2  + 1)}} + \frac{{{{\left( {\sqrt 2  + 1} \right)}^2}}}{{\sqrt 2 (\sqrt 2  + 1)}}\)

\( = \frac{{\sqrt 2  - 1 - 2 + 2 + 2\sqrt 2  + 1}}{{\sqrt 2 (\sqrt 2  + 1)}} = \frac{{3\sqrt 2 }}{{\sqrt 2 (\sqrt 2  + 1)}} = \frac{3}{{\sqrt 2  + 1}} = \frac{{3\left( {\sqrt 2  - 1} \right)}}{{2 - 1}} = 3\left( {\sqrt 2  - 1} \right)\)

b). Ta có: \(\frac{{2 + \sqrt 5 }}{{\sqrt 2  + \sqrt {3 + \sqrt 5 } }} + \frac{{2 - \sqrt 5 }}{{\sqrt 2  - \sqrt {3 - \sqrt 5 } }} = \frac{{2\sqrt 2  + \sqrt {10} }}{{2 + \sqrt {6 + 2\sqrt 5 } }} + \frac{{2\sqrt 2  - \sqrt {10} }}{{2 - \sqrt {6 - 2\sqrt 5 } }}\)

\( = \frac{{2\sqrt 2  + \sqrt {10} }}{{2 + \sqrt {{{\left( {\sqrt 5  + 1} \right)}^2}} }} + \frac{{2\sqrt 2  - \sqrt {10} }}{{2 - \sqrt {{{\left( {\sqrt 5  - 1} \right)}^2}} }}\) \( = \frac{{2\sqrt 2  + \sqrt {10} }}{{2 + \sqrt 5  + 1}} + \frac{{2\sqrt 2  - \sqrt {10} }}{{2 - \sqrt 5  + 1}} = \frac{{2\sqrt 2  + \sqrt {10} }}{{3 + \sqrt 5 }} + \frac{{2\sqrt 2  - \sqrt {10} }}{{3 - \sqrt 5 }}\)

\( = \frac{{\left( {2\sqrt 2  + \sqrt {10} } \right)\left( {3 - \sqrt 5 } \right) + \left( {2\sqrt 2  - \sqrt {10} } \right)\left( {3 + \sqrt 5 } \right)}}{{\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)}}\)

\( = \frac{{6\sqrt 2  - 2\sqrt {10}  + 3\sqrt {10}  - \sqrt {50}  + 6\sqrt 2  + 2\sqrt {10}  - 3\sqrt {10}  - \sqrt {50} }}{{9 - 5}}\)

\( = \frac{{12\sqrt 2  - 2\sqrt {50} }}{4} = \frac{{12\sqrt 2  - 10\sqrt 2 }}{4} = \frac{{2\sqrt 2 }}{4} = \frac{{\sqrt 2 }}{2}.\)