Đáp án:
$A=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}.$
Giải thích các bước giải:
$A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}-\dfrac{1}{\sqrt{x}-2}(x \ge 0; x \ne 4)\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x-2\sqrt{x}+3\sqrt{x}-6}-\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{\sqrt{x}(\sqrt{x}-2)+3(\sqrt{x}-2)}-\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{(\sqrt{x}+3)(\sqrt{x}-2)}-\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}+3)(\sqrt{x}-2)}-\dfrac{5}{(\sqrt{x}+3)(\sqrt{x}-2)}-\dfrac{\sqrt{x}+3}{(\sqrt{x}-2)(\sqrt{x}+3)}\\ =\dfrac{x-4-5-(\sqrt{x}+3)}{(\sqrt{x}-2)(\sqrt{x}+3)}\\ =\dfrac{x-\sqrt{x}-12}{(\sqrt{x}-2)(\sqrt{x}+3)}\\ =\dfrac{x-4\sqrt{x}+3\sqrt{x}-12}{(\sqrt{x}-2)(\sqrt{x}+3)}\\ =\dfrac{\sqrt{x}(\sqrt{x}-4)+3(\sqrt{x}-4)}{(\sqrt{x}-2)(\sqrt{x}+3)}\\ =\dfrac{(\sqrt{x}+3)(\sqrt{x}-4)}{(\sqrt{x}-2)(\sqrt{x}+3)}\\ =\dfrac{\sqrt{x}-4}{\sqrt{x}-2}.$
