Ta có:

+) `6 + 2(sqrt(6) + sqrt(3)+sqrt(2))`

`= 1^2 + (sqrt(2))^2 + (sqrt(3))^2 + 2.sqrt(2).sqrt(3) + 2.1.sqrt(3) + 2.1.sqrt(2)`

`= [(sqrt(2))^2+2sqrt(2).sqrt(3) + (sqrt(3))^2] + (2 . 1 . sqrt(2) + 2 . 1 . sqrt(3)) + 1`

`= (sqrt(2) + sqrt(3)^2 + 2 (sqrt(2)+sqrt(3)) . 1 + 1`

`= (1 + sqrt(2) + sqrt(3))^2`

Suy ra: `sqrt(6 + 2(sqrt(6) + sqrt(3)+sqrt(2))) = sqrt( (1 + sqrt(2) + sqrt(3))^2) = 1 + sqrt(2) + sqrt(3)`

+) `6 + 2(sqrt(6) - sqrt(3)+sqrt(2))`

`= 1^2 + (sqrt(2))^2 + (sqrt(3))^2 - 2.sqrt(2).sqrt(3) + 2.1.sqrt(3) - 2.1.sqrt(2)`

`= [(sqrt(2))^2-2sqrt(2).sqrt(3) + (sqrt(3))^2] + (2 . 1 . sqrt(3) - 2 . 1 . sqrt(2)) + 1`

`= (sqrt(3)-sqrt(2))^2 + 2 (sqrt(3) - sqrt(2)) + 1`

`= (sqrt(3) - sqrt(2) + 1)^2`

Suy ra: `sqrt(6 + 2(sqrt(6) - sqrt(3)+sqrt(2))) = sqrt((sqrt(3)-sqrt(2)+1)^2) = sqrt(3) - sqrt(2) + 1`

Khi đó:

`C = ((1 + sqrt(2) + sqrt(3)) - (sqrt(3)-sqrt(2) + 1))/(sqrt(2))`

`C = (2sqrt(2))/(sqrt(2))`

`C = 2`