Question

rút gọn bt

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

Answer 1

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

Answer 2

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

= \(\frac{\left(2\sqrt{2-\sqrt{3}}+\sqrt{6-3\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2+\sqrt{3}}\right)^2}\) = \(\frac{2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\sqrt{\left(6-3\sqrt{3}\right)\left(2+\sqrt{3}\right)}}{\left(\sqrt{2+\sqrt{3}}\right)^2}\)

= \(\frac{2\sqrt{4-3}+\sqrt{12+6\sqrt{3}-6\sqrt{3}-9}}{2+\sqrt{3}}\) = \(\frac{2\sqrt{1}+\sqrt{3}}{2+\sqrt{3}}\) = \(\frac{2+\sqrt{3}}{2+\sqrt{3}}=1\)

Answer 3

Ta có :

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

\(=\sqrt{4-3}=1\)