Đáp án:
\(\begin{array}{l}
A = \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
B = \dfrac{{\sqrt x + 1}}{{2\sqrt x }}\\
C = \dfrac{{2\sqrt x - 5}}{{\sqrt x - 1}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 1\\
A = \dfrac{{\sqrt x + 1 + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{2\sqrt x + 1}}\\
= \dfrac{{2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{2\sqrt x + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
DK:x > 0\\
B = \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x + 1}}{2}\\
= \dfrac{{\sqrt x + 1}}{{2\sqrt x }}\\
DK:x \ge 0;x \ne 1\\
C = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} + \left( {\sqrt x - 5} \right)\left( {\sqrt x - 1} \right) - 3\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1 + x - 6\sqrt x + 5 - 3\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2x - 7\sqrt x + 5}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {2\sqrt x - 5} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2\sqrt x - 5}}{{\sqrt x - 1}}
\end{array}\)
