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 $x=36$

$\to B=\dfrac{\sqrt{36}+2}{\sqrt{36}-2}=\dfrac{6+2}{6-2}=2$

$A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\,\,\,(x\ge0; x\ne4)\\=\left(\dfrac{\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{1}{\sqrt{x}-2}\right).\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}+2}\\=\dfrac{\sqrt{x}+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}+2}\\=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}$

$C=B.(A-2)\\=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\\=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}\\=\dfrac{-2}{\sqrt{x}-2}$

Đáp án:

Giải thích các bước giải: