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