Question

Rút gọn : \(\frac{4}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}-3}\)

Answer 1

\(\frac{4}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}-3}=\frac{4}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}-\frac{6}{3-\sqrt{3}}=\frac{4\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{6\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\frac{4\sqrt{3}-4}{3-1}+\frac{\sqrt{3}+2}{3-4}-\frac{18+6\sqrt{3}}{9-3}=\frac{2\left(2\sqrt{3}-2\right)}{2}-\left(\sqrt{3}+2\right)-\frac{6\left(3+\sqrt{3}\right)}{6}=2\sqrt{3}-2-\sqrt{3}-2-3-\sqrt{3}=-7\)