Đáp án:
b) \(0 \le a < 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a \ge 0;a \ne 1\\
P = \left[ {\dfrac{{a + 1 + \sqrt a }}{{a + 1}}} \right]:\left[ {\dfrac{1}{{\sqrt a - 1}} - \dfrac{{2\sqrt a }}{{a\left( {\sqrt a - 1} \right) + \left( {\sqrt a - 1} \right)}}} \right]\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}:\dfrac{{a + 1 - 2\sqrt a }}{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}.\dfrac{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}{{{{\left( {\sqrt a - 1} \right)}^2}}}\\
= \dfrac{{a + \sqrt a + 1}}{{\sqrt a - 1}}\\
b)P < 1\\
\to \dfrac{{a + \sqrt a + 1}}{{\sqrt a - 1}} < 1\\
\to \dfrac{{a + \sqrt a + 1 - \sqrt a + 1}}{{\sqrt a - 1}} < 0\\
\to \dfrac{{a + 2}}{{\sqrt a - 1}} < 0\\
\to \sqrt a - 1 < 0\left( {do:a + 2 > 0\forall a \ge 0} \right)\\
\to a < 1\\
\to 0 \le a < 1\\
c)Thay:a = 19 - 8\sqrt 3 \\
= 16 - 2.4.\sqrt 3 + 3 = {\left( {4 - \sqrt 3 } \right)^2}\\
\to P = \dfrac{{19 - 8\sqrt 3 + \sqrt {{{\left( {4 - \sqrt 3 } \right)}^2}} + 1}}{{\sqrt {{{\left( {4 - \sqrt 3 } \right)}^2}} - 1}}\\
= \dfrac{{19 - 8\sqrt 3 + 4 - \sqrt 3 + 1}}{{4 - \sqrt 3 - 1}} = \dfrac{{15 - \sqrt 3 }}{2}
\end{array}\)
