Đáp án+Giải thích các bước giải:
`Q=(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1})(x+\sqrt{x})(x>0,x\ne1)`
`=(\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)})(x+\sqrt{x})`
`=(\frac{(\sqrt{x}+2)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}).\sqrt{x}(\sqrt{x}+1)`
`=\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{(\sqrt{x}+1)^2(\sqrt{x}-1)}).\sqrt{x}(\sqrt{x}+1)`
`=\frac{2\sqrt{x}}{(\sqrt{x}+1)^2(\sqrt{x}-1)}).\sqrt{x}(\sqrt{x}+1)`
`=\frac{2.x(\sqrt{x}+1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}`
`=\frac{2x}{x-1}`
`b)\frac{2x}{x-1}=\frac{2x-2+2}{x-1}=2+\frac{2]{x-1}`
`Q\inZZ`
`=>2\vdots x-1`
`<=>x-1\in{+-1;+-2}`
`->x\in{-1+1;1+1;-2+1;2+1}`
`=>x\in{0;2;-1;3}`
Mà `x>0`
`=>x\in{2;3}` thì `Q\inZZ`
