Đáp án:
`B=1`
`C=4`
Giải thích các bước giải:
`B=({2+\sqrt{2}}/{\sqrt{2}+1}+1)({2-\sqrt{2}}/{\sqrt{2}-1}-1)`
`=({\sqrt{2}.(\sqrt{2}+1)}/{\sqrt{2}+1}+1).({\sqrt{2}.(\sqrt{2}-1)}/{\sqrt{2}-1}-1)`
`=(\sqrt{2}+1).(\sqrt{2}-1)`
`=(\sqrt{2})^2-1^2=2-1=1`
Vậy `B=1`
$\\$
`C=(\sqrt{10}-\sqrt{2}).\sqrt{3+\sqrt{5}}`
`=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}`
`=(\sqrt{5}-1).\sqrt{6+2\sqrt{5}}`
`=(\sqrt{5}-1).\sqrt{5+2\sqrt{5}.1+1^2}`
`=(\sqrt{5}-1).\sqrt{(\sqrt{5}+1)^2}`
`=(\sqrt{5}-1).|\sqrt{5}+1|`
`=(\sqrt{5}-1)(\sqrt{5}+1)`
`=(\sqrt{5})^2-1^2=5-1=4`
Vậy `C=4`
