1)

Với `x > 0, x ne 1`, ta có:

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

`A = (xsqrt(x) (sqrt(x) - 1) + 1 (sqrt(x) - 1))/(sqrt(x) . (1 - x)) + ((sqrt(x) - 1)(x+sqrt(x) + 1))/(sqrt(x) . (sqrt(x) - 1)) + (x+1)/(sqrt(x))`

`A = ((xsqrt(x) + 1)(sqrt(x) - 1))/(-sqrt(x) . (sqrt(x) - 1)(sqrt(x) + 1)) + (x+sqrt(x) + 1)/(sqrt(x)) + (x+1)/(sqrt(x) + 1)`

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

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

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

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

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

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

`A = (sqrt(x) + 1/(sqrt(x))) + 2 >= 2 . sqrt(sqrt(x) . 1/(sqrt(x))) + 2 = 2 . sqrt(1) + 2 = 2 + 2 = 4`

Dấu "=" xảy ra khi và chỉ khi `sqrt(x) = 1/(sqrt(x))`, suy ra `x = 1` (ktm)

Khi đó: `A > 4` (đpcm)

2)

Ta có: `B = 6/A = 6 . 1/A < 6 . 1/4 = 3/2`

Mà `6 > 0, A > 0` nên `B = 6/A > 0`

Khi đó: `0 < B < 3/2`

Mà `B` nhận giá trị nguyên nên `B = 1`

Khi đó: `B = 6/A = 1`

`A = 6`

`(x+2sqrt(x)+1)/(sqrt(x)) = 6`

`x + 2sqrt(x) + 1 = 6sqrt(x)`

`x - 4sqrt(x) + 1 = 0`

`(x - 4sqrt(x) + 4) - 3 = 0`

`(sqrt(x) - 2)^2 - (sqrt(3))^2 = 0`

`(sqrt(x) - 2 - sqrt(3)) (sqrt(x) - 2 + sqrt(3)) = 0`

`sqrt(x) - 2 - sqrt(3) = 0` hoặc `sqrt(x) - 2 + sqrt(3) = 0`

`sqrt(x) = 2 + sqrt(3)` hoặc `sqrt(x) = 2 - sqrt(3)`

`x = 7 + 4sqrt(3)` (tm) hoặc `x = 7-4sqrt(3)` (tm)

Vậy, `x = 7+4sqrt(3)` và `x = 7 - 4sqrt(3)`