a)$\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}$ b)(4+$\sqrt{15}$).($\sqrt{10}$ -$\sqrt{6}$). $\sqrt{4-\sqrt{15}}$ c)$\sqrt{13-\sqrt{160}}$ - $\sqrt{53-4\sqrt{49}}$

`a)` `sqrt(6 - 2sqrt(sqrt(2) + sqrt(12) + sqrt(18) - sqrt(128)))`

Ta có: `sqrt(2) + sqrt(12) + sqrt(18) - sqrt(128)`

`= sqrt(2) + 2sqrt(3) + 3sqrt(2) - 8sqrt(2)`

`= 2sqrt(3) - 4sqrt(2) = sqrt(12) - sqrt(32)`

Vì `12 < 32` nên `sqrt(12) - sqrt(32) < 0`

`->` biểu thức đã cho không xác định

`b)` `(4+sqrt(15))(sqrt(10)-sqrt(6))sqrt(4-sqrt(15))`

`= (4+sqrt(15)) * (sqrt(5)-sqrt(3))^2`

`= (4+sqrt(15)) * (8 - 2sqrt(15))`

`= (4+sqrt(15)) * 2(4 - sqrt(15))`

`= 2(4^2 - (sqrt(15))^2)`

`= 2(16 - 15) = 2`

`c)` `sqrt(13 - sqrt(160)) - sqrt(53 - 4sqrt(49))`

`= sqrt(13 - sqrt(16*10)) - sqrt(53 - 4*7)`

`= sqrt(13 - 4sqrt(10)) - sqrt(53 - 28)`

`= sqrt(13 - 2sqrt(40)) - sqrt(25)`

`= sqrt((sqrt(8) - sqrt(5))^2) - 5`

`= |sqrt(8) - sqrt(5)| - 5`

`= sqrt(8) - sqrt(5) - 5`

`= 2sqrt(2) - sqrt(5) - 5`