Đáp án + Giải thích các bước giải:
$a,ĐKXĐ:$ $\sqrt{x}-2\neq0⇔\sqrt{x}\neq2⇔x\neq4$
$c,\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-2}+\dfrac{6}{x-\sqrt{x}-2}$
$=\dfrac{\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}+1)(\sqrt{x}-2)}-\dfrac{2(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-2)}+\dfrac{6}{(\sqrt{x}+1)(\sqrt{x}-2)}$
$=\dfrac{x-2\sqrt{x}-(2\sqrt{x}+2)+6}{(\sqrt{x}+1)(\sqrt{x}-2)}$
$=\dfrac{x-2\sqrt{x}-2\sqrt{x}-2+6}{(\sqrt{x}+1)(\sqrt{x}-2)}$
$=\dfrac{x-4\sqrt{x}+4}{(\sqrt{x}+1)(\sqrt{x}-2)}$
$=\dfrac{(\sqrt{x}-2)^{2}}{(\sqrt{x}+1)(\sqrt{x}-2)}$
$=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}$
