Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
B = \dfrac{2}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x }} - \dfrac{{3\sqrt x - 5}}{{\sqrt x - x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.} \right)\\
= \dfrac{2}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x }} - \dfrac{{3\sqrt x - 5}}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
= \dfrac{2}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x }} + \dfrac{{3\sqrt x - 5}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2\sqrt x - 1.\left( {\sqrt x - 1} \right) + 3\sqrt x - 5}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2\sqrt x - \sqrt x + 1 + 3\sqrt x - 5}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{4\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{4}{{\sqrt x }}\\
3,\\
a,\\
A = \left( {\dfrac{{\sqrt x - 1}}{{x - \sqrt x }} - \dfrac{{\sqrt x }}{{x + \sqrt x }}} \right):\left( {1 - \dfrac{1}{{\sqrt x }}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.} \right)\\
= \left( {\dfrac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right):\dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
= \left( {\dfrac{1}{{\sqrt x }} - \dfrac{1}{{\sqrt x + 1}}} \right):\dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right) - \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{1}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{x - 1}}\\
b,\\
x = 9 \Rightarrow A = \dfrac{1}{8}
\end{array}\)
Em viết lại đề của câu 2 nhé!
