Đáp án:

\(\begin{array}{l}
a,\\
Q =  - \dfrac{{a - 1}}{{\sqrt a }}\\
b,\\
a > 1\\
c,\\
a = 3 + 2\sqrt 2 \\
d,\\
T < 1,\,\,\,\forall a > 0,a \ne 1
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
Q = {\left[ {\dfrac{{\sqrt a }}{2} - \dfrac{1}{{2\sqrt a }}} \right]^2}.\left[ {\dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}} - \dfrac{{\sqrt a  + 1}}{{\sqrt a  - 1}}} \right]\\
 = {\left[ {\dfrac{{\sqrt a .\sqrt a  - 1}}{{2\sqrt a }}} \right]^2}.\left[ {\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2} - {{\left( {\sqrt a  + 1} \right)}^2}}}{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right)}}} \right]\\
 = {\left( {\dfrac{{a - 1}}{{2\sqrt a }}} \right)^2}.\dfrac{{\left( {a - 2\sqrt a  + 1} \right) - \left( {a + 2\sqrt a  + 1} \right)}}{{{{\sqrt a }^2} - 1}}\\
 = {\left( {\dfrac{{a - 1}}{{2\sqrt a }}} \right)^2}.\dfrac{{a - 2\sqrt a  + 1 - a - 2\sqrt a  - 1}}{{a - 1}}\\
 = {\left( {\dfrac{{a - 1}}{{2\sqrt a }}} \right)^2}.\dfrac{{ - 4\sqrt a }}{{a - 1}}\\
 = \dfrac{{a - 1}}{{2\sqrt a }}.\dfrac{{ - 2}}{1}\\
 =  - \dfrac{{a - 1}}{{\sqrt a }}\\
b,\\
Q < 0\\
 \Leftrightarrow  - \dfrac{{a - 1}}{{\sqrt a }} < 0\\
 \Leftrightarrow \dfrac{{a - 1}}{{\sqrt a }} > 0\\
\sqrt a  > 0,\,\,\,\forall a > 0,a \ne 1\\
 \Rightarrow a - 1 > 0\\
 \Rightarrow a > 1\\
c,\\
Q =  - 2\\
 \Leftrightarrow  - \dfrac{{a - 1}}{{\sqrt a }} =  - 2\\
 \Leftrightarrow \dfrac{{a - 1}}{{\sqrt a }} = 2\\
 \Leftrightarrow a - 1 = 2\sqrt a \\
 \Leftrightarrow a - 2\sqrt a  - 1 = 0\\
 \Leftrightarrow \left( {a - 2\sqrt a  + 1} \right) - 2 = 0\\
 \Leftrightarrow {\left( {\sqrt a  - 1} \right)^2} = 2\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt a  - 1 = \sqrt 2 \\
\sqrt a  - 1 =  - \sqrt 2 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt a  = \sqrt 2  + 1\\
\sqrt a  = 1 - \sqrt 2 
\end{array} \right.\\
\sqrt a  > 0 \Rightarrow \sqrt a  = \sqrt 2  + 1 \Rightarrow a = 3 + 2\sqrt 2 \\
d,\\
T = Q.\sqrt a  = \left( { - \dfrac{{a - 1}}{{\sqrt a }}} \right).\sqrt a  =  - \left( {a - 1} \right) = 1 - a\\
a > 0,a \ne 1 \Rightarrow 1 - a < 1\\
 \Rightarrow T < 1,\,\,\,\forall a > 0,a \ne 1
\end{array}\)