`a = ((\sqrt{(1 + x)^3} + \sqrt{(1 - x)^3})\sqrt{1 - \sqrt{1 - x^2}})/(2 - \sqrt{1 - x^2})` (ĐK: `-1 <= x <= 1`)

`= ((\sqrt{1 + x} + \sqrt{1 - x})[(1 + x) - \sqrt{(1 + x)(1 - x)} + (1 - x)]\sqrt{1 - \sqrt{1 - x^2}})/(2 - \sqrt{1 - x^2})`

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

`= (\sqrt{1 + x} + \sqrt{1 - x})\sqrt{1 - \sqrt{1 - x^2}}`

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

`= (\sqrt{1 + x} + \sqrt{1 - x})(\sqrt{(1 + x) - 2\sqrt{(1 + x)(1 - x)} + (1 - x)})/(\sqrt{2})`

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

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

TH1: Với `0 <= x <= 1 -> \sqrt{1 + x} >= \sqrt{1 - x}`

`= (\sqrt{1 + x} + \sqrt{1 - x})(\sqrt{1 + x} - \sqrt{1 - x})/(\sqrt{2})`

`= ((1 + x) - (1 - x))/(\sqrt{2})`

`= (2x)/(\sqrt{2}) = x\sqrt{2}`

TH2: Với `-1 <= x < 0 -> \sqrt{1 + x} < \sqrt{1 - x}`

`= (\sqrt{1 + x} + \sqrt{1 - x})(\sqrt{1 - x} - \sqrt{1 + x})/(\sqrt{2})`

`= ((1 - x) - (1 + x))/(\sqrt{2})`

`= (-2x)/(\sqrt{2}) = -x\sqrt{2}`

Vậy `a = |x|\sqrt{2}`

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