Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;y \ge 0;y \ne 1\\
A = \dfrac{{{{\left( {2 + \sqrt x } \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2}}}{{2\sqrt x + 3}} + \dfrac{{x + \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{4 + 4\sqrt x + x - \left( {x - 2\sqrt x + 1} \right)}}{{2\sqrt x + 3}} + \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{4 + 4\sqrt x + x - x + 2\sqrt x - 1}}{{2\sqrt x + 3}} + \sqrt x \\
= \dfrac{{6\sqrt x + 3}}{{2\sqrt x + 3}} + \dfrac{{2x + 3\sqrt x }}{{2\sqrt x + 3}}\\
= \dfrac{{2x + 9\sqrt x + 3}}{{2\sqrt x + 3}}\\
B = \dfrac{{{{\left( {\sqrt y + 1} \right)}^2} - 4\sqrt y }}{{\sqrt y - 1}}\\
= \dfrac{{y - 2\sqrt y + 1}}{{\sqrt y - 1}}\\
= \sqrt y - 1\\
b)A - B\\
= \dfrac{{2x + 9\sqrt x + 3}}{{2\sqrt x + 3}} - \sqrt y + 1\\
Khi:x = 6 - 2\sqrt 5 = {\left( {\sqrt 5 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 5 - 1\\
y = 6 + 2\sqrt 5 \Rightarrow \sqrt y = \sqrt 5 + 1\\
\Rightarrow A - B\\
= \dfrac{{2x + 9\sqrt x + 3}}{{2\sqrt x + 3}} - \sqrt y + 1\\
= \dfrac{{2\left( {6 - 2\sqrt 5 } \right) + 9\left( {\sqrt 5 - 1} \right) + 3}}{{2\left( {\sqrt 5 - 1} \right) + 3}} - \left( {\sqrt 5 + 1} \right) + 1\\
= \dfrac{{6 + 5\sqrt 5 }}{{1 + 2\sqrt 5 }} - \sqrt 5 \\
= \dfrac{{6 + 5\sqrt 5 - \sqrt 5 - 10}}{{1 + 2\sqrt 5 }}\\
= \dfrac{{4\sqrt 5 - 4}}{{1 + 2\sqrt 5 }}\\
= \dfrac{{44 - 12\sqrt 5 }}{{19}}
\end{array}$
