`a)`

`B` có nghĩa khi:

`{(x>=0),(x - 5sqrt(x) + 6 != 0),(sqrt(x) - 2 != 0),(3 - sqrt(x) != 0):}`

Vì `x - 5sqrt(x) + 6 = (sqrt(x) - 2)(sqrt(x) - 3)=-(sqrt(x) - 2)(3-sqrt(x))` nên:

`{(x>=0),(sqrt(x) - 2 != 0),(3 - sqrt(x) != 0):}`

`{(x>=0),(x!=4),(x!=9):}`

`b)`

`B = (2sqrt(x) - 9)/((sqrt(x) - 2)(sqrt(x) - 3)) - (sqrt(x) + 3)/(sqrt(x) - 2) + (2sqrt(x) + 1)/(sqrt(x) - 3)`

`B = (2sqrt(x) - 9 - (sqrt(x) + 3)(sqrt(x) - 3) + (2sqrt(x) + 1)(sqrt(x) - 2)) / ((sqrt(x) - 2)(sqrt(x) - 3))`

`B = (2sqrt(x) - 9 - (x - 9) + (2x - 4sqrt(x) + sqrt(x) - 2)) / ((sqrt(x) - 2)(sqrt(x) - 3))`

`B = (2sqrt(x) - 9 - x + 9 + 2x - 3sqrt(x) - 2) / ((sqrt(x) - 2)(sqrt(x) - 3))`

`B = (x - sqrt(x) - 2) / ((sqrt(x) - 2)(sqrt(x) - 3))`

`B = ((sqrt(x) - 2)(sqrt(x) + 1)) / ((sqrt(x) - 2)(sqrt(x) - 3))`

`B = (sqrt(x) + 1) / (sqrt(x) - 3)`

`c)`

`B > 1`

`(sqrt(x) + 1)/(sqrt(x) - 3) > 1`

`(sqrt(x) + 1)/(sqrt(x) - 3) - 1 > 0`

`(sqrt(x) + 1 - (sqrt(x) - 3)) / (sqrt(x) - 3) > 0`

`4 / (sqrt(x) - 3) > 0`

Vì `4>0` nên:

`sqrt(x) - 3 > 0`

`sqrt(x) > 3`

`x > 9`

`d)`

`B = (sqrt(x) + 1)/(sqrt(x) - 3)`

`B= (sqrt(x) - 3 + 4)/(sqrt(x) - 3)`

`B = 1 + 4/(sqrt(x) - 3)`

Để `B` nguyên thì `4 vdots sqrt(x)-3`

Hay `sqrt(x)-3 in U(4)={+-1;+-2;+-4}`

` sqrt(x) - 3 = -4 => sqrt(x) = -1` (loại)

`sqrt(x) - 3 = -2 => sqrt(x) = 1 => x = 1` (thỏa mãn)

`sqrt(x) - 3 = -1 => sqrt(x) = 2 => x = 4` (loại)

`sqrt(x) - 3 = 1 => sqrt(x) = 4 => x = 16` (thỏa mãn)

`sqrt(x) - 3 = 2 => sqrt(x) = 5 => x = 25` (thỏa mãn)

`sqrt(x) - 3 = 4 => sqrt(x) = 7 => x = 49` (thỏa mãn)

Vậy `x in {1, 16, 25, 49}`