Đáp án:
`5)` `A=2`
`6)` `A=2`
Giải thích các bước giải:
`5)` `A=(\sqrt{\sqrt{5}+3}-\sqrt{3-\sqrt{5}})^2`
`A=(\sqrt{\sqrt{5}+3})^2-2.\sqrt{\sqrt{5}+3}. \sqrt{3-\sqrt{5}}+(\sqrt{3-\sqrt{5}})^2`
`A=\sqrt{5}+3-2\sqrt{3^2-5}+3-\sqrt{5}`
`A=\sqrt{5}-\sqrt{5}+3+3-2\sqrt{4}`
`A=6-2.2=2`
Vậy `A=2`
$\\$
`6)` `A=(\sqrt{6}-\sqrt{2}).\sqrt{2+\sqrt{3}}`
`A=(\sqrt{6}-\sqrt{2}).\sqrt{1/ 2. (4+2\sqrt{3})}`
`A=(\sqrt{6}-\sqrt{2}).1/{\sqrt{2}} . \sqrt{ (3+2\sqrt{3}.1+1)}`
`A=\sqrt{2}.(\sqrt{3}-1) . 1/{\sqrt{2}} . \sqrt{ (\sqrt{3}+1)^2}`
`A=(\sqrt{3}-1).(\sqrt{3}+1)`
`A=(\sqrt{3})^2-1^2=2`
Vậy `A=2`
