A = √(-x^2 + 2x + 8) - √(-x^2 + x + 2)
ĐKXĐ: 
{-x^2 + 2x + 8 ≥ 0 <=> {(x + 2)(4 - x) ≥ 0 <=> {-2 ≤ x ≤ 4 <=> -1 ≤ x ≤ 2
{-x^2 + x + 2 ≥ 0          {(2 - x)(x + 1) ≥ 0        {-1 ≤ x ≤ 2
Xét hiệu:
(-x^2 + 2x + 8) - (-x^2 + x + 2) = x + 6 > 0 (vì -1 ≤ x ≤ 2)
=> -x^2 + 2x + 8 > -x^2 + x + 2
=> √(-x^2 + 2x + 8) > √(-x^2 + x + 2)
=> A > 0
Xét: 
A^2 = (x + 2)(4 - x) + (2 - x)(x + 1) - 2√((x + 2)(4 - x)(2 - x)(x + 1))
= (x + 1)(4 - x) + (4 - x) + (x + 2)(2 - x) - (2 - x) - 2√((x + 2)(4 - x)(2 - x)(x + 1))
= (x + 1)(4 - x) + (x + 2)(2 - x) -  2√((x + 2)(4 - x)(2 - x)(x + 1)) + 2
= (√(x + 1)(4 - x) - √(x + 2)(2 - x))^2 + 2
=> A^2 ≥ 2
Mà A > 0
=> A ≥ √2
Dấu "=" xảy ra <=> (√(x + 1)(4 - x) - √(x + 2)(2 - x))^2 = 0 <=> √(x + 1)(4 - x) = √(x + 2)(2 - x)
<=> (x + 1)(4 - x) = (x + 2)(2 - x)
<=> -x^2 + 3x + 4 = -x^2 + 4
<=> 3x = 0
<=> x = 0
Vậy: Min A = √2 <=> x = 0