a) \[K = \frac{1}{{\sqrt {2\sqrt 3 - \sqrt 2 } .\sqrt 2 + \sqrt 3 }}\]

\[K = \frac{1}{{\sqrt {2\sqrt 6 - 2} + 3}}\]

\[K = \frac{{1\left( {\sqrt {2\sqrt 6 - 2} - \sqrt 3 } \right)}}{{\left( {\sqrt {2\sqrt 6 - 2} + \sqrt 3 } \right)\left( {\sqrt {2\sqrt 6 - 2} - \sqrt 3 } \right)}}\]

\[K = \frac{{\sqrt {2\sqrt 6 - 2} - \sqrt 3 }}{{2\sqrt 6 - 2 - 3}} = \frac{{\sqrt {2\sqrt 6 - 2} - \sqrt 3 }}{{2\sqrt 6 - 5}}\]

\[K = \frac{{\left( {\sqrt {2\sqrt 6 - 2} - \sqrt 3 } \right)\left( {2\sqrt 6 + 5} \right)}}{{\left( {2\sqrt 6 - 5} \right)\left( {2\sqrt 6 + 5} \right)}} = \frac{{\left( {\sqrt {2\sqrt 6 - 2} - \sqrt 3 } \right)\left( {2\sqrt 6 + 5} \right)}}{{24 - 25}} = \left( {\sqrt 3 - \sqrt {2\sqrt 6 - 2} } \right)\left( {2\sqrt 6 + 5} \right)\]

b) Ta có: \[\frac{1}{{\sqrt 1 + \sqrt 2 }} = \frac{{1\left( {\sqrt 1 - \sqrt 2 } \right)}}{{\left( {\sqrt 1 + \sqrt 2 } \right)\left( {\sqrt 1 - \sqrt 2 } \right)}} = \frac{{\sqrt 1 - \sqrt 2 }}{{1 - 2}} = \sqrt 2 - 1\]

Tương tự \[\frac{1}{{\sqrt 2 + \sqrt 3 }}\], \[\frac{1}{{\sqrt 3 + \sqrt 4 }}\] …

Và \[\frac{1}{{\sqrt {n - 1} + \sqrt n }} = \frac{{1\left( {\sqrt {n - 1} - \sqrt n } \right)}}{{\left( {\sqrt {n - 1} + \sqrt n } \right)\left( {\sqrt {n - 1} - \sqrt n } \right)}} = \frac{{\sqrt {n - 1} - \sqrt n }}{{n - 1 - n}} = \sqrt n - \sqrt {n - 1} \]

Suy ra \[H = \left( {\sqrt 2 - 1} \right) + \left( {\sqrt 3 - \sqrt 2 } \right) + \left( {2 - \sqrt 3 } \right) + ... + \left( {\sqrt n - \sqrt {n - 1} } \right) = - 1 + \sqrt n = \sqrt n - 1\]

c) Rút gọn biểu thức (P)

\[P = \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right):\left( {\frac{{\sqrt x - 1}}{{\sqrt x }} + \frac{{1 - \sqrt x }}{{x + \sqrt x }}} \right)\] (đk: x > 0)

\[P = \left( {\frac{{x - 1}}{{\sqrt x }}} \right):\left( {\frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x + \sqrt x }} + \frac{{1 - \sqrt x }}{{x + \sqrt x }}} \right)\]

\[P = \left( {\frac{{x - 1}}{{\sqrt x }}} \right):\left( {\frac{{x - \sqrt x }}{{x + \sqrt x }}} \right)\]

\[P = \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x }}.\frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} = \frac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }}\]

Với \[x = \frac{2}{{2 + \sqrt 3 }} = \frac{{2\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}} = \frac{{4 - 2\sqrt 3 }}{{4 - 3}} = 4 - 2\sqrt 3 \] khi đó P có giá trị là:

\[P = \frac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }} = \frac{{{{\left( {\sqrt {4 - 2\sqrt 3 } + 1} \right)}^2}}}{{\sqrt {4 - 2\sqrt 3 } }} = \frac{{{{\left( {\sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} + 1} \right)}^2}}}{{\sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} }} = \frac{{{{\left( {\left| {1 - \sqrt 3 } \right| + 1} \right)}^2}}}{{\left| {1 - \sqrt 3 } \right|}} = \frac{{{{\left( {\sqrt 3 - 1 + 1} \right)}^2}}}{{\sqrt 3 - 1}} = \frac{3}{{\sqrt 3 - 1}}\]

Cho tam giác ABC vuông tại B. Biết AB = 3cm, BC = 4cm. Số đo góc A?