Đáp án:
\(\begin{array}{l}
a)\dfrac{2}{3}\\
b)\dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
c)0 < x < 4;x \ne 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 16\\
\to B = \dfrac{{\sqrt {16} - 2}}{{\sqrt {16} - 1}} = \dfrac{{4 - 2}}{{4 - 1}} = \dfrac{2}{3}\\
b)A = \dfrac{{x - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
c)A.B < 0\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x - 2}}{{\sqrt x - 1}} < 0\\
\to \dfrac{{\sqrt x - 2}}{{\sqrt x }} < 0\\
\to \sqrt x - 2 < 0\left( {do:\sqrt x > 0\forall x > 0} \right)\\
\to 0 < x < 4;x \ne 1
\end{array}\)
