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