\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)\(=\left(3+\sqrt{2}\right)-\left(3-\sqrt{2}\right)=2\sqrt{2}\)\(\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(3-a\right)^2}\)\(=\left|a+3\right|+\left|3-a\right|=3+a+3-a=6\)Câu trả lời: \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)\(=\left(3+\sqrt{2}\right)-\left(3-\sqrt{2}\right)=2\sqrt{2}\)\(\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(3-a\right)^2}\)\(=\left|a+3\right|+\left|3-a\right|=3+a+3-a=6\)