a)(2+√3)√(7-4√3) b) √(2-√3) .(√6 +√2)

$\displaystyle \begin{array}{{>{\displaystyle}l}} a) \ \left( 2+\sqrt{3}\right)\sqrt{7-4\sqrt{3}}\\ =\left( 2+\sqrt{3}\right)\sqrt{4-2.2\sqrt{3} +3}\\ =\left( 2+\sqrt{3}\right)\sqrt{\left( 2-\sqrt{3}\right)^{2}}\\ =\left( 2+\sqrt{3}\right) |2-\sqrt{3} |\\ =\left( 2+\sqrt{3}\right)\left( 2-\sqrt{3}\right) \ \left( vì\ 2 >\sqrt{3}\right)\\ =4-3=1\\ b) \ \sqrt{2-\sqrt{3}} .\left(\sqrt{6} +\sqrt{2}\right)\\ =\sqrt{2-\sqrt{3}} .\sqrt{2}\left(\sqrt{3} +1\right)\\ =\sqrt{4-2\sqrt{3}} .\left(\sqrt{3} +1\right)\\ =\sqrt{1-2\sqrt{3} +3} .\left(\sqrt{3} +1\right)\\ =\sqrt{\left( 1-\sqrt{3}\right)^{2}} .\left(\sqrt{3} +1\right)\\ =|1-\sqrt{3} |.\left(\sqrt{3} +1\right)\\ =\left(\sqrt{3} -1\right)\left(\sqrt{3} +1\right) =3-1=2\ \left( vì\ \sqrt{3} >1\right) \ \\ \end{array}$

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`