Question

Tính giá trị của biểu thức:

a, A = \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)}^2\)

b, B = \(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)

Answer 1

`A=sqrt{(2-sqrt5)^2}+sqrt{(2sqrt2-sqrt5)^2}`

`A=|2-sqrt5|+|2sqrt2-sqrt5|`

`A=\sqrt5-2+2sqrt2-sqrt5`

`A=2sqrt2-2`

`b)B=sqrt{(sqrt7-2sqrt2)^2}+sqrt{(3-2sqrt2)^2}`

`B=|sqrt7-2sqrt2|+|3-2sqrt2|`

`A=2sqrt2-sqrt7+3-2sqrt2`

`A=3-sqrt7`

Answer 2

a,=> A=\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\sqrt{2}\right)^2}=2-\sqrt{5}+\sqrt{5}-2\sqrt{2}=2-2\sqrt{2}\)

b tương tự

Answer 3

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

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

\(=\sqrt{2}\)

b) Ta có: \(B=\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)

\(=2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\)

\(=3-\sqrt{7}\)