Rút gọn biểu thức: \(\left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{1 - a}}} \right)^2} = 1\)

* Hướng dẫn giải

 \(\left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{1 - a}}} \right)^2}\)

\(= \left[ {\dfrac{{1 - {{\left( {\sqrt a } \right)}^3}}}{{1 - \sqrt a }} + \sqrt a } \right]{\left[ {\dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}} \right]^2}\)

\(= \left[ {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}{{1 - \sqrt a }} + \sqrt a } \right]{\left[ {\dfrac{1}{{1 + \sqrt a }}} \right]^2}\)

\( = \left( {1 + 2\sqrt a + a} \right) \cdot {\left( {\dfrac{1}{{1 + \sqrt a }}} \right)^2}\)

\(= \dfrac{{{{\left( {1 + \sqrt a } \right)}^2}}}{{{{\left( {1 + \sqrt a } \right)}^2}}}\)

= 1