`\qquad Q=((\sqrt{x}+1)/(\sqrt{x}-2)-(2\sqrt{x})/(\sqrt{x}+2)+(5\sqrt{x}+2)/(4-x)):(3\sqrt{x}-x)/(x+4\sqrt{x}+4)`
`a)` ĐKXĐ: `{(x>=0),(\sqrt{x}-2\ne0),(\sqrt{x}+2\ne0 AAx),(4-x\ne0),(x+4\sqrt{x}+4\ne0),(3\sqrt{x}-x\ne0):}<=>{(x>=0),(x\ne4),(x\ne0),(x\ne9):}<=>{(x>0),(x\ne4),(x\ne9):}`
Với `x>0;x\ne4;x\ne9` thì
`Q=((\sqrt{x}+1)(\sqrt{x}+2)-2\sqrt{x}(\sqrt{x}-2)-5\sqrt{x}-2)/((\sqrt{x}-2)(\sqrt{x}+2)).(\sqrt{x}+2)^2/(\sqrt{x}(3-\sqrt{x}))`
`Q=(x+2\sqrt{x}+\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2)/(\sqrt{x}-2). (\sqrt{x}+2)/(\sqrt{x}(3-\sqrt{x}))`
`Q=(-x+2\sqrt{x})/(\sqrt{x}-2). (\sqrt{x}+2)/(\sqrt{x}(3-\sqrt{x}))`
`Q=(-\sqrt{x}(\sqrt{x}-2))/(\sqrt{x}-2). (\sqrt{x}+2)/(\sqrt{x}(3-\sqrt{x}))`
`Q=(\sqrt{x}+2)/(\sqrt{x}-3)`
Vậy `Q=(\sqrt{x}+2)/(\sqrt{x}-3)` với `x>0;x\ne4;x\ne9`
`b) Q=(\sqrt{x}+2)/(\sqrt{x}-3)=2`
`<=> \sqrt{x}+2=2\sqrt{x}-6`
`<=> \sqrt{x}=8`
`<=> x=64(\text{tm})`
Vậy `x=64` thì `Q=2`
`c) Q=(\sqrt{x}+2)/(\sqrt{x}-3)<0`
Do `\sqrt{x}+2>0`
`=> \sqrt{x}-3<0`
`<=> \sqrt{x}<3`
`<=> 0<=x<9`
Vậy `0<=x<9` và `x\ne4` thì `Q<0`
