ĐKXĐ: `x > 0, x != 9`

`a)` Thay `x = 36` (tmđk) vào biểu thức `text{A}`, ta được:

`text{A} = (sqrt(36) + 2) / (1 + sqrt(36)) = (6 + 2) / (1 + 6) = 8/7`

Vậy tại `x = 36` thì `text{A} = 8/7`

`b)text{B} = ((2sqrt(x)) / (x - sqrt(x) - 6) + sqrt(x) / (sqrt(x) - 3)) : sqrt(x) / (sqrt(x) - 3)`

`= [(2sqrt(x)) / ((sqrt(x) - 3)(sqrt(x) + 2)) + sqrt(x) / (sqrt(x) - 3)] * (sqrt(x) - 3) / sqrt(x)`

`= (2sqrt(x) + sqrt(x)(sqrt(x) + 2)) / ((sqrt(x) - 3)(sqrt(x) + 2)) * (sqrt(x) - 3) / sqrt(x)`

`= (2sqrt(x) + x + 2sqrt(x)) / ((sqrt(x) - 3)(sqrt(x) + 2)) * (sqrt(x) - 3) / sqrt(x)`

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

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

`= (sqrt(x) + 4) / (sqrt(x) + 2)`

Vậy `text{B} = (sqrt(x) + 4) / (sqrt(x) + 2)`

`c)` Ta có:

`text{P = A} * text{B}= (sqrt(x) + 2) / (sqrt(x) + 1) * (sqrt(x) + 4) / (sqrt(x) + 2)`

`= (sqrt(x) + 4) / (sqrt(x) + 1)`

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

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

Vì `x > 0` nên `3 / (sqrt(x) + 1)` đạt giá trị lớn nhất khi `x` nhỏ nhất

Vì `x in ZZ`, `x > 0, x != 9` nên `x` nhỏ nhất là `x = 1`

Khi `x = 1`, ta có:

`text{P} = 1 + 3 / (sqrt(1) + 1) = 1 + 3 / 2 = 5 / 2`

Vậy `text{P}_(text{max}) = 5 / 2` khi `x = 1`