Đáp án:
c) \(\left[ \begin{array}{l}
a = 9\\
a = 4\\
a = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a \ge 0;a \ne 1\\
P = \dfrac{{3a + 3\sqrt a - 3 - \left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right) + \sqrt a - 1}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{3a + 3\sqrt a - 3 - a + 4 + \sqrt a - 1}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{2a + 4\sqrt a }}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{2\sqrt a }}{{\sqrt a - 1}}\\
b)\left| P \right| = 1\\
\to \left[ \begin{array}{l}
P = 1\\
P = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{2\sqrt a }}{{\sqrt a - 1}} = 1\\
\dfrac{{2\sqrt a }}{{\sqrt a - 1}} = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\sqrt a = \sqrt a - 1\\
2\sqrt a = - \sqrt a + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt a = - 1\left( l \right)\\
3\sqrt a = 1
\end{array} \right.\\
\to \sqrt a = \dfrac{1}{3}\\
\to a = \dfrac{1}{9}\\
c)P = \dfrac{{2\sqrt a }}{{\sqrt a - 1}} = \dfrac{{2\left( {\sqrt a - 1} \right) + 2}}{{\sqrt a - 1}}\\
= 2 + \dfrac{2}{{\sqrt a - 1}}\\
P \in N\\
\Leftrightarrow \dfrac{2}{{\sqrt a - 1}} \in N\\
\to \sqrt a - 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt a - 1 = 2\\
\sqrt a - 1 = - 2\left( l \right)\\
\sqrt a - 1 = 1\\
\sqrt a - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
a = 9\\
a = 4\\
a = 0
\end{array} \right.
\end{array}\)
