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