Question

Rút gọn \(\frac{\sqrt{3}-1}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{1-\frac{\sqrt{3}}{2}}} \)
Giúp mình vs ạ, mk cần gấp gấp lắm ạ

Answer 1

\(=\frac{2\left(\sqrt{3}-1\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\left(\sqrt{3}+1\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{2\left(\sqrt{3}-1\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2\left(\sqrt{3}+1\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\frac{2\left(\sqrt{3}-1\right)}{2+\sqrt{3}+1}+\frac{2\left(\sqrt{3}+1\right)}{2-\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{3+\sqrt{3}}+\frac{2\left(\sqrt{3}+1\right)}{3-\sqrt{3}}\)

\(=\frac{2\left(\sqrt{3}-1\right)\left(3-\sqrt{3}\right)+2\left(\sqrt{3}+1\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\frac{16\sqrt{3}}{6}=\frac{8\sqrt{3}}{3}\)