Chứng minh biểu thức sau không phụ thuộc vào biến \(a\):

                                             \(\left( {\frac{1}{{2 + 2\sqrt a }} + \frac{1}{{2 - 2\sqrt a }} - \frac{{{a^2} + 1}}{{1 - {a^2}}}} \right).\left( {1 + \frac{1}{a}} \right)\)    với \(a > 0,a \ne 1\).

        ta có: \(\frac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a  = \frac{{1 - {{\left( {\sqrt a } \right)}^3}}}{{1 - \sqrt a }} + \sqrt a  = 1 + \sqrt a  + a + \sqrt a \)\( = 1 + 2\sqrt a  + a = {\left( {1 + \sqrt a } \right)^2}\)

và \(\frac{{1 - \sqrt a }}{{1 - a}} = \frac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}} = \frac{1}{{1 + \sqrt a }}\)

Do đó: \({\rm{VT}} = \left( {\frac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\frac{{1 - \sqrt a }}{{1 - a}}} \right)^2} = {\left( {1 + \sqrt a } \right)^2} \cdot \frac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}} = 1 = {\rm{VP}}\) (đpcm).

b) Ta có: \({\rm{VT}} = \frac{{a + b}}{{{b^2}}}\sqrt {\frac{{{a^2}{b^4}}}{{{a^2} + 2ab + {b^2}}}}  = \frac{{a + b}}{{{b^2}}}\sqrt {\frac{{{a^2}{b^4}}}{{{{\left( {a + b} \right)}^2}}}}  = \frac{{a + b}}{{{b^2}}}.\frac{{\left| a \right|.{b^2}}}{{a + b}} = \left| a \right| = {\rm{VP}}\) (đpcm).

(do \(a + b > 0\) và \(b \ne 0\)).

         ta có: \({\rm{VT}} = \frac{3}{2}\sqrt 6  + 2\sqrt {\frac{2}{3}}  - 4\sqrt {\frac{3}{2}}  = \frac{3}{2}\sqrt 6  + 2\sqrt {\frac{6}{9}}  - 4\sqrt {\frac{6}{4}} \)

\( = \frac{3}{2}\sqrt 6  + \frac{2}{3}\sqrt 6  - 2\sqrt 6  = \left( {\frac{3}{2} + \frac{2}{3} - 2} \right)\sqrt 6  = \frac{{\sqrt 6 }}{6} = {\rm{VP}}\) (đpcm).

b) b) Ta có \({\rm{VT}} = \left( {x\sqrt {\frac{6}{x}}  + \sqrt {\frac{{2x}}{3}}  + \sqrt {6x} } \right):\sqrt {6x}  = \left( {\sqrt {6x}  + \frac{{\sqrt {6x} }}{3} + \sqrt {6x} } \right):\sqrt {6x} {\rm{ }}\)

\( = \frac{7}{3}\sqrt {6x} :\sqrt {6x}  = \frac{7}{3} = 2\frac{1}{3} = {\rm{VP}}\) (đpcm).