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