Đáp án:
d. a=6
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a > 0;a \ne \left\{ {1;2} \right\}\\
b.D = \left( {\dfrac{{a\sqrt a - 1}}{{a - \sqrt a }} - \dfrac{{a\sqrt a + 1}}{{a + \sqrt a }}} \right):\dfrac{{a + 2}}{{a - 2}}\\
= \left[ {\dfrac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \dfrac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right].\dfrac{{a - 2}}{{a + 2}}\\
= \left( {\dfrac{{a + \sqrt a + 1 - a + \sqrt a - 1}}{{\sqrt a }}} \right).\dfrac{{a - 2}}{{a + 2}}\\
= \dfrac{{2\sqrt a }}{{\sqrt a }}.\dfrac{{a - 2}}{{a + 2}}\\
= \dfrac{{2a - 4}}{{a + 2}}\\
c.Thay:a = 3 - 2\sqrt 2 \\
\to D = \dfrac{{2\left( {3 - 2\sqrt 2 } \right) - 4}}{{3 - 2\sqrt 2 + 2}} = \dfrac{{6 - 4\sqrt 2 - 4}}{{5 - 2\sqrt 2 }} = \dfrac{{2 - 4\sqrt 2 }}{{5 - 2\sqrt 2 }}\\
d.D = \dfrac{{2a - 4}}{{a + 2}} = \dfrac{{2\left( {a + 2} \right) - 8}}{{a + 2}}\\
= 2 - \dfrac{8}{{a + 2}}\\
D \in Z\\
\to \dfrac{8}{{a + 2}} \in Z\\
\Leftrightarrow a + 2 \in U\left( 8 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
a + 2 = 8\\
a + 2 = - 8\\
a + 2 = 4\\
a + 2 = - 4\\
a + 2 = 2\\
a + 2 = - 2\\
a + 2 = 1\\
a + 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
a = 6\\
a = - 10\left( l \right)\\
a = 2\left( l \right)\\
a = - 6\left( l \right)\\
a = 0\left( l \right)\\
a = - 4\left( l \right)\\
a = - 1\left( l \right)\\
a = - 3\left( l \right)
\end{array} \right.\\
\to a = 6
\end{array}\)
