Đáp án:
b. a=6
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a > 0;a \ne \left\{ {1;2} \right\}\\
A = \left( {\dfrac{{a\sqrt a - 1}}{{a - \sqrt a }} - \dfrac{{a\sqrt a + 1}}{{a + \sqrt a }}} \right):\dfrac{{a + 2}}{{a - 2}}\\
= \left[ {\dfrac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \dfrac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right].\dfrac{{a - 2}}{{a + 2}}\\
= \left( {\dfrac{{a + \sqrt a + 1 - a + \sqrt a - 1}}{{\sqrt a }}} \right).\dfrac{{a - 2}}{{a + 2}}\\
= \dfrac{{2\sqrt a }}{{\sqrt a }}.\dfrac{{a - 2}}{{a + 2}}\\
= \dfrac{{2a - 4}}{{a + 2}}\\
b.A = 1\\
\to \dfrac{{2a - 4}}{{a + 2}} = 1\\
\to 2a - 4 = a + 2\\
\to a = 6\left( {TM} \right)
\end{array}\)
