Đáp án:
$\begin{array}{l}
1)a)A = 2\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = 2\\
\Leftrightarrow \sqrt x + 1 = 2\sqrt x - 2\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow x = 9\left( {tmdk} \right)\\
Vậy\,x = 9\\
b)A.B\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}.\left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} - \dfrac{{\sqrt x - 1}}{{x - 1}}} \right)\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right) - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x - 1}}.\dfrac{{x - 2\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= 1
\end{array}$
Vậy A.B không phụ thuộc vào x
$\begin{array}{l}
c)A \ge B\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} \ge \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} \ge 0\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} \ge 0\\
\Leftrightarrow \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2}}}{{x - 1}} \ge 0\\
\Leftrightarrow \dfrac{{4\sqrt x }}{{x - 1}} \ge 0\\
\Leftrightarrow x - 1 > 0\\
\Leftrightarrow x > 1\\
Vậy\,x > 1
\end{array}$
Câu 2 chép lại đề bài cho thật rõ nhé
