Điều kiện xác định: `-1 <= x <= 1`

`A = (sqrt(1-sqrt(1-x^2)) * (sqrt((1+x)^3) + sqrt((1-x)^3))) / (2 - sqrt(1-x^2))`

Ta có:

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

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

`sqrt(1-sqrt(1-x^2)) = sqrt((2 - 2sqrt(1-x^2))/2)`

`= sqrt(((1+x) - 2sqrt(1+x)sqrt(1-x) + (1-x))/2)`

`= sqrt((sqrt(1+x) - sqrt(1-x))^2 / 2)`

`= (|sqrt(1+x) - sqrt(1-x)|)/sqrt(2)`

Suy ra:

`A = ((|sqrt(1+x) - sqrt(1-x)|)/sqrt(2) * (sqrt(1+x) + sqrt(1-x))(2 - sqrt(1-x^2))) / (2 - sqrt(1-x^2))`

Vì `-1 <= x <= 1` nên `0 <= sqrt(1-x^2) <= 1`

`=> 2 - sqrt(1-x^2) > 0`

`A = (|sqrt(1+x) - sqrt(1-x)|(sqrt(1+x) + sqrt(1-x)))/sqrt(2)`

`A = |(sqrt(1+x))^2 - (sqrt(1-x))^2|/sqrt(2)`

`A = |(1+x) - (1-x)|/sqrt(2)`

`A = |2x|/sqrt(2)`

`A = sqrt(2)|x|`