Đáp án:
$B=2$ khi $x=36$
$A=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}$
$C=\dfrac{-2}{\sqrt{x}-2}$
Giải thích các bước giải:
$B=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\,\,\,(x\ge0; x\ne4)$
Khi $x=36$
$\to B=\dfrac{\sqrt{36}+2}{\sqrt{36}-2}=\dfrac{6+2}{6-2}=2$
$A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\,\,\,(x\ge0; x\ne4)\\=\left(\dfrac{\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{1}{\sqrt{x}-2}\right).\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}+2}\\=\dfrac{\sqrt{x}+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}+2}\\=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}$
$C=B.(A-2)\\=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\\=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}\\=\dfrac{-2}{\sqrt{x}-2}$
