\(\sqrt{20}-\sqrt{45}+\sqrt{6+2\sqrt{5}}=\sqrt{2^2.5}-\sqrt{3^2.5}+\sqrt{\left(\sqrt{5}+1\right)^2}=2\sqrt{5}-3\sqrt{5}+\sqrt{5}+1=1\)
\(\sqrt{20}-2-\sqrt{\left(\sqrt{5}-2\right)^2}=2\sqrt{5}-2-\left|\sqrt{5}-2\right|=2\sqrt{5}-2-\sqrt{5}+2=\sqrt{5}\)
\(\left(\sqrt{27}+3\sqrt{12}-2\sqrt{3}\right):\sqrt{3}=\left(3\sqrt{3}+6\sqrt{3}-2\sqrt{3}\right):\sqrt{3}=7\sqrt{3}:\sqrt{3}=7\)
\(\sqrt{50}-3\sqrt{8}+\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{5^2.2}-3\sqrt{2^2.2}+\sqrt{\left(\sqrt{2}+1\right)^2}=5\sqrt{2}-6\sqrt{2}+\sqrt{2}+1=1\)
1) \(A=\sqrt{20}-\sqrt{45}+\sqrt{6+2\sqrt{5}}\)
\(=2\sqrt{5}-3\sqrt{5}+\sqrt{5}+1\)
=1
2) Ta có: \(B=\sqrt{20}-2-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2\sqrt{5}-2-\sqrt{5}+2\)
\(=\sqrt{5}\)
3) Ta có: \(\left(\sqrt{27}+3\sqrt{12}-2\sqrt{3}\right):\sqrt{3}\)
\(=\left(3\sqrt{3}+6\sqrt{3}-2\sqrt{3}\right):\sqrt{3}\)
=7
4) Ta có: \(\sqrt{50}-3\sqrt{8}+\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=5\sqrt{2}-6\sqrt{2}+\sqrt{2}+1\)
=1
