`a)P=(15\sqrt{x}-11)/(x+2\sqrt{x}-3)+(3\sqrt{x}-2)/(1-\sqrt{x})-(2\sqrt{x}+3)/(\sqrt{x}+3)(x>=0,x\ne1)`

`=(15\sqrt{x}-11)/((\sqrt{x}-1)(\sqrt{x}+3))-(3\sqrt{x}-2)/(\sqrt{x}-1)-(2\sqrt{x}+3)/(\sqrt{x}+3)`

`=(15\sqrt{x}-11)/((\sqrt{x}-1)(\sqrt{x}+3))-((3\sqrt{x}-2)(\sqrt{x}+3))/((\sqrt{x}-1)(\sqrt{x}+3))-((2\sqrt{x}+3)(\sqrt{x}-1))/((\sqrt{x}-1)(\sqrt{x}+3))`

`=(15\sqrt{x}-11-(3x+9\sqrt{x}-2\sqrt{x}-6)-(2x-2\sqrt{x}+3\sqrt{x}-3))/((\sqrt{x}+3)(\sqrt{x}-1))`

`=(15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3)/((\sqrt{x}-1)(\sqrt{x}+3))`

`=(-5x+7\sqrt{x}-2)/((\sqrt{x}-1)(\sqrt{x}+3))`

`=-(5x-7\sqrt{x}+2)/((\sqrt{x}-1)(\sqrt{x}+3))`

`=-((\sqrt{x}-1)(5\sqrt{x}-2))/((\sqrt{x}-1)(\sqrt{x}+3))`

`=-(5\sqrt{x}-2)/(\sqrt{x}+3)`

`b)P=-(5(\sqrt{x}+3)-17)/(\sqrt{x}+3)`

`=-5+17/(\sqrt{x}+3)`

Ta có:

`\sqrt{x}+3>=3`

`->17/(\sqrt{x}+3)<=17/3`

`->P=-5+17/(\sqrt{x}+3)<=-5+17/3=2/3`

Dấu "=" xảy ra: `x=0(N)`

Vậy: `...`