`c)`

`B=(\sqrt{x})/(\sqrt{x}+2)-(4\sqrt{x})/(4-x)`

`B=(\sqrt{x}(\sqrt{x}-2))/((\sqrt{x}+2)(\sqrt{x}-2))+(4\sqrt{x})/((\sqrt{x}+2)(\sqrt{x}-2))`

`B=(x-2\sqrt{x}+4\sqrt{x})/((\sqrt{x}+2)(\sqrt{x}-2))`

`B=(x+2\sqrt{x})/((\sqrt{x}+2)(\sqrt{x}-2))`

`B=(\sqrt{x}(\sqrt{x}+2))/((\sqrt{x}+2)(\sqrt{x}-2))`

`B=(\sqrt{x})/(\sqrt{x}-2)`

`P=A·B`

`P=(x-3)/(\sqrt{x})·(\sqrt{x})/(\sqrt{x}-2)`

`P=(x-3)/(\sqrt{x}-2)`

Để `P≤6` thì

`(x-3)/(\sqrt{x}-2)≤6`

`(x-3)/(\sqrt{x}-2)-6≤0`

`(x-3-6(\sqrt{x}-2))/(\sqrt{x}-2)≤0`

`(x-6\sqrt{x}+9)/(\sqrt{x}-2)≤0`

`((\sqrt{x}-3)^2)/(\sqrt{x}-2)≤0`

Ta có: `(\sqrt{x}-3)^2≥0`

`⇒\sqrt{x}-2≤0`

`⇒\sqrt{x}≤2`

`⇒x≤4`

`⇒0<x<4`

Vậy `x in {1;2;3}`