Đáp án:
a. \(\dfrac{{a + 1}}{{1 + \sqrt a + a}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a \ge 0;a \ne \left\{ {\dfrac{1}{4};1} \right\}\\
P = 1 + \left[ {\dfrac{{\left( {2\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\left( {1 - \sqrt a } \right)\left( {\sqrt a + 1} \right)}} - \dfrac{{2a\sqrt a + a - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}} \right].\dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\\
= 1 + \left[ {\dfrac{{2\sqrt a - 1}}{{1 - \sqrt a }} - \dfrac{{2a\sqrt a + a - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}} \right].\dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\\
= 1 + \left[ {\dfrac{{\left( {2\sqrt a - 1} \right)\left( {1 + \sqrt a + a} \right) - 2a\sqrt a - a + \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}} \right].\dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\\
= 1 + \left[ {\dfrac{{2\sqrt a - 1 + 2a - \sqrt a + 2a\sqrt a - a - 2a\sqrt a - a + \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}} \right].\dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\\
= 1 + \dfrac{{2\sqrt a - 1}}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}.\dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\\
= 1 + \dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}\\
= 1 - \dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {1 + \sqrt a + a} \right)}}\\
= 1 - \dfrac{{\sqrt a }}{{1 + \sqrt a + a}}\\
= \dfrac{{1 + \sqrt a + a - \sqrt a }}{{1 + \sqrt a + a}}\\
= \dfrac{{a + 1}}{{1 + \sqrt a + a}}\\
b.P = \dfrac{{\sqrt 6 }}{{1 + \sqrt 6 }}\\
\to \dfrac{{a + 1}}{{1 + \sqrt a + a}} = \dfrac{{\sqrt 6 }}{{1 + \sqrt 6 }}\\
\to a + 1 + a\sqrt 6 + \sqrt 6 = \sqrt 6 + \sqrt {6a} + a\sqrt 6 \\
\to a - \sqrt {6a} + 1 = 0\\
\to \left[ \begin{array}{l}
\sqrt a = \dfrac{{\sqrt 6 + \sqrt 2 }}{2}\\
\sqrt a = \dfrac{{\sqrt 6 - \sqrt 2 }}{2}
\end{array} \right. \to \left[ \begin{array}{l}
a = 2 + \sqrt 3 \\
a = 2 - \sqrt 3
\end{array} \right.\\
c.P > \dfrac{2}{3}\\
\to \dfrac{{a + 1}}{{1 + \sqrt a + a}} > \dfrac{2}{3}\\
\to \dfrac{{3a + 3 - 2 - 2\sqrt a - 2a}}{{3\left( {1 + \sqrt a + a} \right)}} > 0\\
\to a - 2\sqrt a + 1 > 0\left( {do:1 + \sqrt a + a > 0\forall a \ge 0} \right)\\
\to {\left( {\sqrt a - 1} \right)^2} > 0\left( {ld} \right)\forall a \ge 0;a \ne 1\\
\to dpcm
\end{array}\)
