$\text{ĐKXĐ: $x\neq4, x>0$}$
$\text{B=$\frac{3}{\sqrt{x}+2}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{2x-3\sqrt{x}+6}{x-4}$}$
$\text{=$\frac{3}{\sqrt{x}+2}-\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{2x-3\sqrt{x}+6}{(\sqrt{x}+2)(\sqrt{x}-2)}$}$
$\text{=$\frac{3(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}-\frac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-2)}+\frac{2x-3\sqrt{x}+6}{(\sqrt{x}+2)(\sqrt{x}-2)}$}$
$\text{=$\frac{3\sqrt{x}-6}{(\sqrt{x}+2)(\sqrt{x}-2)}-\frac{x+2\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}+\frac{2x-3\sqrt{x}+6}{(\sqrt{x}+2)(\sqrt{x}-2)}$}$
$\text{=$\frac{3\sqrt{x}-6-(x+2\sqrt{x})+2x-3\sqrt{x}+6}{(\sqrt{x}+2)(\sqrt{x}-2)}$}$
$\text{=$\frac{3\sqrt{x}-6-x-2\sqrt{x}+2x-3\sqrt{x}+6}{(\sqrt{x}+2)(\sqrt{x}-2)}$}$
$\text{=$\frac{x-2\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}$}$
$\text{=$\frac{\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}$}$
$\text{=$\frac{\sqrt{x}}{\sqrt{x}+2}$}$
$\text{Vậy với $x\neq4, x>0$ thì B=$\frac{\sqrt{x}}{\sqrt{x}+2}$}$
