Đáp án:

Giải thích các bước giải:

`a)` Với `x = 25` (tmdk) ta có:

`A = (\sqrt25 - 4)/\sqrt25 = (5 - 4)/5 = 1/5`

`b)` Với `x > 0 ; x ne 9` ta có:

`B = 4/(\sqrtx - 3) + (x - 7\sqrtx - 12)/(x - 9)`

`= (4(\sqrtx + 3) + x - 7\sqrtx - 12)/((\sqrtx - 3)(\sqrtx + 3))`

`= (4\sqrtx + 12 + x - 7\sqrtx - 12)/((\sqrtx - 3)(\sqrtx + 3))`

`= (x - 3\sqrtx)/((\sqrtx - 3)(\sqrtx + 3))`

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

`= \sqrtx/(\sqrtx + 3)`

`c) P = A . B = (\sqrtx - 4)/\sqrtx . \sqrtx/(\sqrtx + 3)`

`= (\sqrtx - 4)/(\sqrtx + 3)`

`= (\sqrtx + 3 - 7)/(\sqrtx + 3)`

`= 1 - 7/(\sqrtx + 3)`

Do `-7/(\sqrtx + 3) > 0` nên `P < 1`

Lại có: `\sqrtx + 3 >= 3` nên `-7/(\sqrtx + 3) >= -7/3`

`=> P >= 1 + (-7)/3 = -4/3`

Khi đó: `-4/3 < P < 1`

`TH1 : P = -1`

`=> (\sqrtx - 4)/(\sqrtx + 3) = -1`

`=> \sqrtx - 4 = -3 - \sqrtx`

`=> 2\sqrtx = 1`

`=> \sqrtx = 1/2`

`=> x = 1/4` (tmdk)         

`TH2 : P = 0` khi đó: `\sqrtx - 4 = 0`

`=> \sqrtx = 4`

`=> x = 16`

            Vậy `x \in {1/4 ; 16}` thì `P` nguyên