A=(√a/2-1/2√a)^2*(√a-1/√a+1-√a+1/√a-1) Rút gọn,tìm giá trị để A =-2

Đáp án:

$\begin{array}{l}
Dkxd:a > 0;a \ne 1\\
A = {\left( {\dfrac{{\sqrt a }}{2} - \dfrac{1}{{2\sqrt a }}} \right)^2}.\left( {\dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}} - \dfrac{{\sqrt a  + 1}}{{\sqrt a  - 1}}} \right)\\
 = {\left( {\dfrac{{a - 1}}{{2\sqrt a }}} \right)^2}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2} - {{\left( {\sqrt a  + 1} \right)}^2}}}{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right)}}\\
 = \dfrac{{{{\left( {a - 1} \right)}^2}}}{{4a}}.\dfrac{{a - 2\sqrt a  + 1 - a - 2\sqrt a  + 1}}{{a - 1}}\\
 = \dfrac{{a - 1}}{{4a}}.\left( { - 4\sqrt a } \right)\\
 = \dfrac{{1 - a}}{{\sqrt a }}\\
A =  - 2\\
 \Rightarrow \dfrac{{1 - a}}{{\sqrt a }} =  - 2\\
 \Rightarrow 1 - a =  - 2\sqrt a \\
 \Rightarrow a - 2\sqrt a  = 1\\
 \Rightarrow a - 2\sqrt a  + 1 = 2\\
 \Rightarrow {\left( {\sqrt a  - 1} \right)^2} = 2\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt a  = 1 + \sqrt 2 \\
\sqrt a  = 1 - \sqrt 2 \left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow a = {\left( {1 + \sqrt 2 } \right)^2} = 3 + 2\sqrt 2 \left( {tmdk} \right)
\end{array}$