Đáp án:
b) \(\dfrac{{a\sqrt a - 1}}{{{{\left( {\sqrt a + 1} \right)}^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a > 0;a \ne 1\\
b)A = \dfrac{{a + \sqrt a + 1}}{{a + 1}}:\left[ {\dfrac{1}{{\sqrt a - 1}} + \dfrac{{2\sqrt a }}{{a\left( {\sqrt a - 1} \right) + \left( {\sqrt a - 1} \right)}}} \right]\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}:\left[ {\dfrac{{a + 1 + 2\sqrt a }}{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}} \right]\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}:\dfrac{{{{\left( {\sqrt a + 1} \right)}^2}}}{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}.\dfrac{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
= \dfrac{{a\sqrt a - 1}}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
c)Thay:a = 4 + 2\sqrt 3 = 3 + 2\sqrt 3 .1 + 1\\
= {\left( {\sqrt 3 + 1} \right)^2}\\
\to A = \dfrac{{\left( {4 + 2\sqrt 3 } \right)\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - 1}}{{{{\left( {\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + 1} \right)}^2}}}\\
= \dfrac{{\left( {4 + 2\sqrt 3 } \right)\left( {\sqrt 3 + 1} \right) - 1}}{{{{\left( {\sqrt 3 + 1 + 1} \right)}^2}}}\\
= \dfrac{{10 + 6\sqrt 3 - 1}}{{7 + 4\sqrt 3 }} = \dfrac{{9 + 6\sqrt 3 }}{{7 + 4\sqrt 3 }}\\
d)A > 1\\
\to \dfrac{{a\sqrt a - 1}}{{{{\left( {\sqrt a + 1} \right)}^2}}} > 1\\
\to \dfrac{{a\sqrt a - 1 - a - 2\sqrt a - 1}}{{{{\left( {\sqrt a + 1} \right)}^2}}} > 0\\
\to a\sqrt a - a - 2\sqrt a - 2 > 0\left( {do:{{\left( {\sqrt a + 1} \right)}^2} > 0\forall a \ge 0} \right)\\
\to \sqrt a > 2,269530842\\
\to a > 5,150770243
\end{array}\)
