Đáp án:
$A=\sqrt{x}$
Giải thích các bước giải:
Sửa $\sqrt3$ thành $\sqrt{x}$ nhé.
$A=(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}):\dfrac{2\sqrt{x}}{x-9}$
$=[\dfrac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}]:\dfrac{2\sqrt{x}}{x-9}$
$=\dfrac{x-3\sqrt{x}+x+3\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}:\dfrac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$=\dfrac{2x}{(\sqrt{x}-3)(\sqrt{x}+3)}\times \dfrac{(\sqrt{x}-3)(\sqrt{x}+3)}{2\sqrt{x}}$
$=\sqrt{x}$
