\(B=\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}\) ( Hình như đề này mới đúng =)))
\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{2}{3-1}=\dfrac{2}{2}=1\)
B=\(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3+1}}=\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{4}}=\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{2}\)
B=\(\dfrac{2}{2\left(\sqrt{3}-1\right)}-\dfrac{\sqrt{3}-1}{2\left(\sqrt{3}-1\right)}=\dfrac{2-\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}\)
B=\(\dfrac{2-\sqrt{3}+1}{2\left(\sqrt{3}-1\right)}=\dfrac{3-\sqrt{3}}{2\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}\)
B=\(\dfrac{\sqrt{3}}{2}\)