Question

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

 

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

 

Answer 1

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

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

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

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

Answer 2

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

b: \(=3-2\sqrt{2}+3+2\sqrt{2}=6\)

Answer 3

`a, sqrt((sqrt 5 - sqrt 2)^2) - sqrt((sqrt 5 + sqrt 2)^2)`

`= sqrt 5 - sqrt 2 - sqrt 5 - sqrt 2`

`= - 2 sqrt 2`

`b, = 2 sqrt 2 - 3 + 2 sqrt 2 + 3`

`= 4 sqrt 2`