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