Ta có: \(A=\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{3}+3}+\sqrt{12-2\cdot2\sqrt{3}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(=3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)
Ta có:
\(B=5\cdot\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\dfrac{3}{2}}\right)^2\)
\(=5\left[\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{5}\right)\right]^2+\left[\dfrac{1}{\sqrt{2}}\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}-\sqrt{3}\right)\right]^2\)
\(=\dfrac{5}{2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\right)^2+\dfrac{1}{2}\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{3}\right)^2\)
\(=\dfrac{5}{2}\left(\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}\right)^2+\dfrac{1}{2}\left(\sqrt{3}-1+\sqrt{5}+1-\sqrt{3}\right)^2\)
\(=\dfrac{5}{2}\left(\sqrt{3}\right)^2+\dfrac{1}{2}\left(\sqrt{5}\right)^2=\dfrac{5}{2}\cdot3+\dfrac{1}{2}\cdot5=\dfrac{15}{2}+\dfrac{5}{2}=\dfrac{20}{2}=10\)
giúp e vs ạ ;-;
giúp e vs ạ :<