Đáp án:
`c) =1`
`b)=1`
Giải thích các bước giải:
`a) (2+\sqrt{3})\sqrt{7-4\sqrt{3}}`
`=(2+\sqrt{3})\sqrt{(\sqrt{4}-4\sqrt{3}+\sqrt{3}}`
`=(2+\sqrt{3})\sqrt{(\sqrt{4}-\sqrt{3})^2}`
`=(2+\sqrt{3})|\sqrt{4}-\sqrt{3}|`
`=(2+\sqrt{3})(\sqrt{4}-\sqrt{3})`
`=2\sqrt{4}-2\sqrt{3}+\sqrt{12}-\sqrt{9}`
`=4-2\sqrt{3}+2\sqrt{3}-3`
`=1`
`b) \sqrt{2-\sqrt{3}}(\sqrt{6}+\sqrt{2})`
`=\sqrt{(4-2\sqrt{3})/2}(\sqrt{6}+\sqrt{2})`
`=\sqrt{((\sqrt{3})^2-2\sqrt{3}+1)/2}(\sqrt{6}+\sqrt{2})`
`=\sqrt{((\sqrt{3}-1)^2)/2}(\sqrt{6}+\sqrt{2})`
`=|(\sqrt{3}-1)/(\sqrt{2})|(\sqrt{6}+\sqrt{2})`
`=(\sqrt{3}-1)/(\sqrt{2}). \sqrt{2}(\sqrt{3}+1)`
`=1`
