Đáp án:
b. \(x > 0;x \ne 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 + 1}} - \sqrt {5 - 2\sqrt 5 .1 + 1} \\
= \sqrt 5 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \sqrt 5 - \sqrt 5 + 1 = 1\\
DK:x > 0;x \ne 1\\
B = \left[ {\dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right]:\dfrac{{\sqrt x }}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{{x - 1}}{x}\\
b.B < A\\
\to \dfrac{{x - 1}}{x} < 1\\
\to \dfrac{{x - 1 - x}}{x} < 0\\
\to - \dfrac{1}{x} < 0\\
\to x > 0;x \ne 1
\end{array}\)
