Rút gọn biêu thức: \(A = \left( {{{1 - a\sqrt a } \over {1 - \sqrt a }} + \sqrt a } \right).{\left( {{{1 - \sqrt a } \over {1 - a}}} \right)^2}\)\(\,\,\,\left( {a \ge 0;\,a \ne 1}...

* Hướng dẫn giải

Ta có:

\(A = \left( {{{1 - a\sqrt a } \over {1 - \sqrt a }} + \sqrt a } \right).{\left( {{{1 - \sqrt a } \over {1 - a}}} \right)^2}\)

\(\eqalign{  &  = \left( {{{{1^3} - {{\left( {\sqrt a } \right)}^3}} \over {1 - \sqrt a }} + \sqrt a } \right).{\left( {{{1 - \sqrt a } \over {1 - a}}} \right)^2}  \cr  &  = \left( {{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a  + a} \right)} \over {1 - \sqrt a }} + \sqrt a } \right).{\left( {{{1 - \sqrt a } \over {1 - a}}} \right)^2}  \cr  &  = \left( {1 + 2\sqrt a  + a} \right).{{{{\left( {1 - \sqrt a } \right)}^2}} \over {{{\left( {1 - \sqrt a } \right)}^2}{{\left( {1 + \sqrt a } \right)}^2}}} \cr} \) 

\(\begin{array}{l}
= {\left( {1 + \sqrt a } \right)^2}.\frac{{{{\left( {1 - \sqrt a } \right)}^2}}}{{{{\left( {1 - \sqrt a } \right)}^2}{{\left( {1 + \sqrt a } \right)}^2}}}\\
= 1
\end{array}\)