Question

1) \(\sqrt{7-2\sqrt{10}}\) - \(\sqrt{7+2\sqrt{10}}\)

2) \(\sqrt{4-2\sqrt{3}}\) + \(\sqrt{4+2\sqrt{3}}\)

3) \(\sqrt{6-4\sqrt{2}}\) + \(\sqrt{22-12\sqrt{2}}\)

Answer 1

\(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}=-2\sqrt{2}\)

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

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

\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)

\(=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)

Answer 2

1: Ta có: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)

\(=-2\sqrt{2}\)

2: Ta có: \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

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

\(=2\sqrt{3}\)