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

`đk: x >= 0, x ne 4, x ne 9`

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

`B = (2x + 6sqrtx - x - 9sqrtx)/(x - 9)`

`B = (x - 3sqrtx)/(x - 9)`

`B = (sqrtx(sqrtx - 3))/((sqrtx - 3)(sqrtx + 3))`

`B = sqrtx/(sqrtx + 3)`

`M = sqrtx/(sqrtx - 2) : sqrtx/(sqrtx + 3)`

`M = sqrtx/(sqrtx - 2) . (sqrtx + 3)/sqrtx`

`M = (sqrtx + 3)/(sqrtx - 2) (với x ne 0)`

`M = 1 + 5/(sqrtx - 2)`

Để `M in ZZ` thì `5/(sqrtx - 2) in ZZ`

`to sqrtx - 2 in Ư(5) = {-5, -1, 1, 5}`

`+) sqrtx - 2 = -5 to sqrtx = -3 (loại)`

`+) sqrtx - 2 = -1 to sqrtx = 1 to x = 1 (tmđk)`

`+)sqrtx - 2 = 1 to sqrtx = 3 to x = 9 (loại)`

`+)sqrtx - 2 = 5 to sqrtx = 7 to x = 49 (tmđk)`

Vậy `x in {1, 49}`