Đáp án:
A = $\dfrac{\sqrt{x}}{\sqrt{x}+3}$
Giải thích các bước giải:
A = $\dfrac{\sqrt{x} + 15}{x-9}$ - $\dfrac{x}{x-3\sqrt{x}}$ + $\dfrac{2\sqrt{x}+5}{\sqrt{x}+3}$
= $\dfrac{(\sqrt{x} + 15)\sqrt{x}}{\sqrt{x}(x-9)}$ - $\dfrac{x(\sqrt{x}+3)}{\sqrt{x}(x-9)}$ + $\dfrac{(2\sqrt{x}+5)(\sqrt{x}-3)\sqrt{x}}{\sqrt{x}(x-9)}$
= $\dfrac{x+15\sqrt{x}-x\sqrt{x}-3x+2x\sqrt{x}+5x-6x-15\sqrt{x}}{\sqrt{x}(x-9)}$
= $\dfrac{-3x+x\sqrt{x}}{\sqrt{x}(x-9)}$
= $\dfrac{x(\sqrt{x}-3)}{\sqrt{x}(x-9)}$
= $\dfrac{\sqrt{x}}{\sqrt{x}+3}$
