Đáp án:
\(\begin{array}{l}
a)\dfrac{2}{{\sqrt a - 1}}\\
b)0 < a < 1\\
c)\sqrt 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{\sqrt a - 1 + \sqrt a + 1}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.\dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
= \dfrac{{2\sqrt a }}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.\dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
= \dfrac{2}{{\sqrt a - 1}}\\
b)A < 0\\
\to \dfrac{2}{{\sqrt a - 1}} < 0\\
\to \sqrt a - 1 < 0\\
\to 0 < a < 1\\
c)Thay:a = 3 + 2\sqrt 2 \\
= 2 + 2\sqrt 2 .1 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to A = \dfrac{2}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - 1}} = \dfrac{2}{{\sqrt 2 + 1 - 1}}\\
= \dfrac{2}{{\sqrt 2 }} = \sqrt 2
\end{array}\)
