Rút gọn K.

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6 \ne 0\\\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  - 3} \right) \ne 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt x  \ne 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\x \ne 9\end{array} \right..\)

\(\begin{array}{l}K = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6}} - \dfrac{{6 - \sqrt {xy} }}{{\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6}}\\ = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)}} - \dfrac{{6 - \sqrt {xy} }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt y  + 2} \right)}}\\ = \dfrac{{\left( {2\sqrt x  + 3\sqrt y } \right)\left( {\sqrt x  + 3} \right) - \left( {6 - \sqrt {xy} } \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + 6\sqrt x  + 3\sqrt {xy}  + 9\sqrt y  - \left( {6\sqrt x  - 18 - x\sqrt y  + 3\sqrt {xy} } \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + x\sqrt y  + 9\sqrt y  + 18}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}} = \dfrac{{\left( {\sqrt y  + 2} \right)\left( {x + 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{x + 9}}{{x - 9}}.\end{array}\)

Vậy \(K = \dfrac{{x + 9}}{{x - 9}}\)  với \(x;y \ge 0;x \ne 9.\)