Giải thích các bước giải:
a/ $R=(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}})(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4})$
$\text{Với $x>0$ và $x \neq 4$}$
⇔ $R=(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}(\sqrt{x}-2)})(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{(\sqrt{x}-2)(\sqrt{x}+2)})$
⇔ $R=\dfrac{\sqrt{x}.\sqrt{x}-4}{\sqrt{x}(\sqrt{x}-2)}.\dfrac{\sqrt{x}-2+4}{(\sqrt{x}-2)(\sqrt{x}+2)}$
⇔ $R=\dfrac{x-4}{\sqrt{x}(\sqrt{x}-2)}.\dfrac{\sqrt{x}+2}{x-4}$
⇔ $R=\dfrac{\sqrt{x}+2}{\sqrt{x}(\sqrt{x}-2)}$
b/ $\text{Tại $x=4+2\sqrt{3}=3+2\sqrt{3}+1=(\sqrt{3}+1)^2$}$
⇔ $R=\dfrac{\sqrt{(\sqrt{3}+1)^2}+2}{\sqrt{(\sqrt{3}+1)^2}(\sqrt{(\sqrt{3}+1)^2}-2)}$
⇔ $R=\dfrac{\sqrt{3}+1+2}{(\sqrt{3}+1)(\sqrt{3}+1-2)}$
⇔ $R=\dfrac{\sqrt{3}+3}{(\sqrt{3}+1)(\sqrt{3}-1)}$
⇔ $R=\dfrac{\sqrt{3}+3}{3-1}$
⇔ $R=\dfrac{\sqrt{3}+3}{2}$
Chúc bạn học tốt !!!
