Đáp án:
a. \(\dfrac{{1 - a}}{{\sqrt a }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a > 0;a \ne 1\\
A = {\left[ {\dfrac{{a - 1}}{{2\sqrt a }}} \right]^2}.\left[ {\dfrac{{{{\left( {\sqrt a - 1} \right)}^2} - {{\left( {\sqrt a + 1} \right)}^2}}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}} \right]\\
= {\left( {\dfrac{{a - 1}}{{2\sqrt a }}} \right)^2}.\left[ {\dfrac{{a - 2\sqrt a + 1 - a - 2\sqrt a - 1}}{{a - 1}}} \right]\\
= {\left( {\dfrac{{a - 1}}{{2\sqrt a }}} \right)^2}.\dfrac{{\left( { - 4\sqrt a } \right)}}{{a - 1}}\\
= \dfrac{{{{\left( {a - 1} \right)}^2}}}{{4a}}.\dfrac{{\left( { - 4\sqrt a } \right)}}{{a - 1}}\\
= \dfrac{{ - \left( {a - 1} \right)}}{{\sqrt a }} = \dfrac{{1 - a}}{{\sqrt a }}\\
b.A < 0\\
\to \dfrac{{1 - a}}{{\sqrt a }} < 0\\
\to 1 - a < 0\left( {do:\sqrt a > 0\forall a > 0} \right)\\
\to 1 < a\\
c.A = - 2\\
\to \dfrac{{1 - a}}{{\sqrt a }} = - 2\\
\to 1 - a = - 2\sqrt a \\
\to a - 2\sqrt a - 1 = 0\\
\to \left[ \begin{array}{l}
\sqrt a = 1 + \sqrt 2 \\
\sqrt a = 1 - \sqrt 2 \left( l \right)
\end{array} \right.\\
\to a = 3 + 2\sqrt 2
\end{array}\)
