\(b,=\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{2\sqrt{3}\left(2+\sqrt{3}\right)}{2+\sqrt{3}}=4+2\sqrt{3}-2\sqrt{3}=4\)
Bài 2:
\(A=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{x+4}=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(x+4\right)}=\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{4+2\sqrt{3}-\left(4\sqrt{3}+6\right)\left(2-\sqrt{3}\right)}{4-3}=4+2\sqrt{3}-8\sqrt{3}+12-12+6\sqrt{3}=4\)
Bài 2
\(A=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(x+4\right)}=\dfrac{1}{\sqrt{x}-2}\)
