Ta có :

A=\(\sqrt{12+6\sqrt{3}}+\sqrt{12-6\sqrt{3}}\)

   =\(\sqrt{9+6\sqrt{3}+3}+\sqrt{9-6\sqrt{3+3}}\)

   =\(\sqrt{3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{3^2-2.3\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

   =\(3+\sqrt{3}+3-\sqrt{3}=6\)

Vậy A =6

1.\(\sqrt{11-4\sqrt{7}}=\sqrt{11-2\cdot2\sqrt{7}}\)=\(\sqrt{7-2\cdot2\cdot\sqrt{7}+4}=\sqrt{\left(\sqrt{7}-2\right)^2}\)=\(\sqrt{7}-2\)

2.\(\left(3-\sqrt{9}\right)\sqrt{11+6\sqrt{6}}=\left(3-3\right)\sqrt{11+6\sqrt{6}}\)=\(0\cdot\sqrt{11+6\sqrt{6}}=0\)

3.\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\)\(\sqrt{15-2\cdot3\cdot\sqrt{6}}+\sqrt{35-2\cdot2\cdot3\cdot\sqrt{2}\cdot\sqrt{3}}\)

                                                      =\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{35-2\cdot\left(2\sqrt{2}\right)\left(3\sqrt{3}\right)}\)=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{27-2\cdot\left(2\sqrt{2}\right)\left(3\sqrt{3}\right)+8}\)

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

\(\sqrt{12-6\sqrt{3}}=\sqrt{9-6\sqrt{3}+3}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}\)

\(=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=\sqrt{16+8\sqrt{3}+3}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}\)

\(=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=\sqrt{9-6\sqrt{5}+5}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}\)

\(=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{12-6\sqrt{3}}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{12-6\sqrt{3}}=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=3-\sqrt{5}\)

học dốt quá

Cho sửa phần mẫu số của câu trên thành \(\sqrt{6}+\sqrt{2}\)

\(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-|2\sqrt{3}+1|}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{4+2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)

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

\(=\frac{2\sqrt{3+|\sqrt{3}-1|}}{\sqrt{6}+\sqrt{2}}\)

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

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

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

\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)