Đáp án:
\(\begin{array}{l}
A = \dfrac{{x + 8}}{{\sqrt x + 1}}\\
B = \dfrac{{\sqrt x }}{{\sqrt x + 5}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0\\
A = \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 1} \right) - 3\sqrt x + 6}}{{\sqrt x \left( {\sqrt x + 1} \right)}}:\dfrac{{\sqrt x + 1 - \sqrt x }}{{\sqrt x }}\\
= \dfrac{{x + 3\sqrt x + 2 - 3\sqrt x + 6}}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x }}{1}\\
= \dfrac{{x + 8}}{{\sqrt x + 1}}\\
DK:x > 0;x \ne 25\\
B = \dfrac{{\sqrt x - \sqrt x + 5}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}:\dfrac{{\sqrt x - \sqrt x + 5}}{{\sqrt x \left( {\sqrt x - 5} \right)}}\\
= \dfrac{5}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 5} \right)}}{5}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 5}}
\end{array}\)
