Question

Bài 2: 

a) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)                       b) \(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}\)

c) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)                    d) \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)

e) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)                            f) \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)

Hộ mk vs ạ

Answer 1

Answer 2

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

b)\(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}=\left(5-2\sqrt{6}\right)-\left(5+2\sqrt{6}\right)=5-2\sqrt{6}-5-2\sqrt{6}\)\(=-4\sqrt{6}\)

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

d)\(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}=\sqrt{2}+1-5+\sqrt{2}=2\sqrt{2}-4\)

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

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

Answer 3

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

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

=6

c) Ta có: \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

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

\(=2\sqrt{5}\)