a/ ĐKXĐ: $x\ne 9,x\ge 0$
$A\,=\dfrac{2\sqrt x}{\sqrt x-3}-\dfrac{\sqrt x}{3-\sqrt x}-\dfrac{3x+3}{x-9}\\\quad=\dfrac{2\sqrt x}{\sqrt x-3}+\dfrac{\sqrt x}{\sqrt x-3}-\dfrac{3x+3}{(\sqrt x-3)(\sqrt x+3)}\\\quad =\dfrac{2\sqrt x+\sqrt x}{\sqrt x-3}-\dfrac{3x+3}{(\sqrt x-3)(\sqrt x+3)}\\\quad=\dfrac{3\sqrt x(\sqrt x+3)}{(\sqrt x+3)(\sqrt x-3)}-\dfrac{3x+3}{(\sqrt x-3)(\sqrt x+3)}\\\quad =\dfrac{3x+9\sqrt x-3x-3}{(\sqrt x-3)(\sqrt x+3)}\\\quad =\dfrac{9\sqrt x-3}{x-9}$
Vậy $A=\dfrac{9\sqrt x-3}{x-9}$ với $x\ge 0,x\ne 9$
b/ Thay $x=49$ (thỏa mãn điều kiện) vào biểu thức $A$
$A\,=\dfrac{9\sqrt{49}-3}{49-9}\\\quad =\dfrac{9.7-3}{40}\\\quad =\dfrac{63-3}{40}\\\quad =\dfrac{60}{40}\\\quad =\dfrac{3}{2}$
Vậy $A=\dfrac{3}{2}$ với $x=49$
