D=($\frac{x+3}{x-9}$ +$\frac{1}{√x+3}$ ) : $\frac{√x}{√x -3}$ a) rút gọn và tìm ĐKXĐ b) tính giá trị D khi x=4-2√3

$a)ĐKXĐ:\left\{\begin{array}{l} x>0\\ x-9 \ne 0 \\ \sqrt{x}+3 \ne 0\\ \sqrt{x}-3 \ne 0\\ \sqrt{x} \ne 0\end{array} \right. <=>x>0,x \ne 9\\ D=\left( \dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3} \right) : \dfrac{\sqrt{x}}{\sqrt{x}-3}\\ =\left( \dfrac{x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)} \right).\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ D=\left( \dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3} \right) : \dfrac{\sqrt{x}}{\sqrt{x}-3}\\ =\left( \dfrac{x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)} \right).\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ =\dfrac{x+\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}.\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ b)x=4-2\sqrt{3}=3-2\sqrt{3}+1=(\sqrt{3}-1)^2\\ D=\dfrac{\sqrt{(\sqrt{3}-1)^2}+1}{\sqrt{(\sqrt{3}-1)^2}+3}\\=\dfrac{\sqrt{3}-1+1}{\sqrt{3}-1+3}\\=\dfrac{\sqrt{3}}{\sqrt{3}+2}$