Đáp án:
\(\begin{array}{l}
a)\dfrac{{\sqrt a }}{{a + \sqrt a + 1}}\\
b)P < \sqrt P \\
c)TH1:\left| P \right| \ge P\\
\to a = 0\\
TH2:\left| P \right| < P\\
\to a \ne 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{2a + 4 + \left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right) - 2\left( {a + \sqrt a + 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}\\
= \dfrac{{2a + 4 + a + \sqrt a - 2 - 2a - 2\sqrt a - 2}}{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}\\
= \dfrac{{a - \sqrt a }}{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}\\
= \dfrac{{\sqrt a }}{{a + \sqrt a + 1}}\\
b)Xét:P - \sqrt P = \sqrt P \left( {\sqrt P - 1} \right)\\
Do:\sqrt P \ge 0\forall a\\
TH1:\sqrt P \ge 1\\
\to \sqrt {\dfrac{{\sqrt a }}{{a + \sqrt a + 1}}} \ge 1\\
\to \dfrac{{\sqrt a }}{{a + \sqrt a + 1}} \ge 1\\
\to \dfrac{{\sqrt a - a - \sqrt a - 1}}{{a + \sqrt a + 1}} \ge 0\\
\to \dfrac{{ - a - 1}}{{a + \sqrt a + 1}} \ge 0\\
\to \dfrac{{a + 1}}{{a + \sqrt a + 1}} \le 0\\
Do:a \ge 0 \to \left\{ \begin{array}{l}
a + 1 > 0\\
a + \sqrt a + 1 > 0
\end{array} \right.\\
\to \dfrac{{a + 1}}{{a + \sqrt a + 1}} \le 0\left( {vôlý} \right)\\
\to TH1\left( {loại} \right)\\
TH2:\sqrt P < 1\\
\to \sqrt {\dfrac{{\sqrt a }}{{a + \sqrt a + 1}}} < 1\\
\to \dfrac{{\sqrt a }}{{a + \sqrt a + 1}} < 1\\
\to \dfrac{{\sqrt a - a - \sqrt a - 1}}{{a + \sqrt a + 1}} < 0\\
\to \dfrac{{ - a - 1}}{{a + \sqrt a + 1}} < 0\\
\to \dfrac{{a + 1}}{{a + \sqrt a + 1}} > 0\\
Do:a \ge 0 \to \left\{ \begin{array}{l}
a + 1 > 0\\
a + \sqrt a + 1 > 0
\end{array} \right.\\
\to \dfrac{{a + 1}}{{a + \sqrt a + 1}} > 0\left( {ld} \right)\\
\to TH2:\left( {TM} \right)\\
KL:P < \sqrt P \\
c)TH1:\left| P \right| \ge P\\
\to \left[ \begin{array}{l}
P \ge P\left( {ld} \right)\\
P \le - P
\end{array} \right.\\
\to 2P \le 0\\
\to P \le 0\\
\to \dfrac{{\sqrt a }}{{a + \sqrt a + 1}} \le 0\\
\Leftrightarrow a = 0\left( {do:a + \sqrt a + 1 > 0\forall a} \right)\\
TH2:\left| P \right| < P\\
\to \left[ \begin{array}{l}
P < P\left( l \right)\\
P > - P
\end{array} \right.\\
\to 2P > 0\\
\to P > 0\\
\to \dfrac{{\sqrt a }}{{a + \sqrt a + 1}} > 0\\
\Leftrightarrow a \ne 0
\end{array}\)
