Tập xác định: $x≥0$ và $x\neq1$
$K=\dfrac{15\sqrt[]{x}-11}{x+2\sqrt[]{x}-3}+\dfrac{3\sqrt[]{x}}{1-\sqrt[]{x}}-\dfrac{2\sqrt[]{x}+3}{\sqrt[]{x}+3}$
$=\dfrac{15\sqrt[]{x}-11}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}-\dfrac{3\sqrt[]{x}}{\sqrt[]{x}-1}-\dfrac{2\sqrt[]{x}+3}{\sqrt[]{x}+3}$
$=\dfrac{15\sqrt[]{x}-11-3\sqrt[]{x}(\sqrt[]{x}+3)-(2\sqrt[]{x}+3)(\sqrt[]{x}-1)}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}$
$=\dfrac{-5x+5\sqrt[]{x}-8}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}$
