Đáp án: -1
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ge 0;y \ge 0;x \ne 9\\
B = \dfrac{{2\sqrt x + 3\sqrt y }}{{\sqrt {xy} + 2\sqrt x - 3\sqrt y - 6}} - \dfrac{{6 - \sqrt {xy} }}{{\sqrt {xy} + 2\sqrt x + 3\sqrt y + 6}}\\
= \dfrac{{2\sqrt x + 3\sqrt y }}{{\sqrt x \left( {\sqrt y + 2} \right) - 3\left( {\sqrt y + 2} \right)}} - \dfrac{{6 - \sqrt {xy} }}{{\sqrt x \left( {\sqrt y + 2} \right) + 3\left( {\sqrt y + 2} \right)}}\\
= \dfrac{{2\sqrt x + 3\sqrt y }}{{\left( {\sqrt y + 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{6 - \sqrt {xy} }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt y + 2} \right)}}\\
= \dfrac{{\left( {2\sqrt x + 3\sqrt y } \right)\left( {\sqrt x + 3} \right) - \left( {6 - \sqrt {xy} } \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt y + 2} \right)\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x + 6\sqrt x + 3\sqrt {xy} + 9\sqrt y - 6\sqrt x + 18 - x\sqrt y - 3\sqrt {xy} }}{{\left( {\sqrt y + 2} \right)\left( {x - 9} \right)}}\\
= \dfrac{{2x + 9\sqrt y - x\sqrt y + 18}}{{\left( {\sqrt y + 2} \right)\left( {x - 9} \right)}}\\
= \dfrac{{\left( {\sqrt y + 2} \right)\left( { - x + 9} \right)}}{{\left( {\sqrt y + 2} \right)\left( {x - 9} \right)}}\\
= - 1
\end{array}$
