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