Đáp án:
$C=\dfrac{3\sqrt{x}}{2\sqrt{x}+2}$ với $x\neq9;x≥0$
Giải thích các bước giải:
$C=\bigg(\dfrac{x+9}{x-9}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\bigg):\dfrac{2\sqrt{x}+2}{x-3\sqrt{x}}$ ĐK: $x\neq9;x≥0$
$C=\dfrac{x+9-\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}\cdot\dfrac{x-3\sqrt{x}}{2\sqrt{x}+2}$
$C=\dfrac{x+9-x+3\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}\cdot\dfrac{\sqrt{x}(\sqrt{x}-3)}{2\sqrt{x}+2}$
$C=\dfrac{3(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}\cdot\dfrac{\sqrt{x}(\sqrt{x}-3)}{2\sqrt{x}+2}$
$C=\dfrac{3\sqrt{x}}{2\sqrt{x}+2}$
Vậy $C=\dfrac{3\sqrt{x}}{2\sqrt{x}+2}$ với $x\neq9;x≥0$
