`a)``ĐK:{(xge0),(xne1):}`
`Q=((x+2)/(x\sqrt{x}-1)+\sqrt{x}/(x+\sqrt{x}+1)+1/(1-\sqrt{x})):(\sqrt{x}-1)/2`
`=((x+2)/((\sqrt{x}-1)(x+\sqrt{x}+1))+\sqrt{x}/(x+\sqrt{x}+1)-1/(\sqrt{x}-1)). 2/(\sqrt{x}-1)`
`=(x+2+\sqrt{x}(\sqrt{x}-1)-(x+\sqrt{x}+1))/((\sqrt{x}-1)(x+\sqrt{x}+1)). 2/(\sqrt{x}-1)`
`=(x+2+x-\sqrt{x}-x-\sqrt{x}-1)/((\sqrt{x}-1)(x+\sqrt{x}+1)).
2/(\sqrt{x}-1)`
`=(x-2\sqrt{x}+1)/((\sqrt{x}-1)(x+\sqrt{x}+1)). 2/(\sqrt{x}-1)`
`=(\sqrt{x}-1)^2/((\sqrt{x}-1)(x+\sqrt{x}+1)). 2/(\sqrt{x}-1)`
`=2/(x+\sqrt{x}+1)`
`b)``2/(x+\sqrt{x}+1)`
`=2/((x-2\sqrt{x}+1)+3\sqrt{x})`
`=2/((\sqrt{x}-1)^2+3\sqrt{x})`
Vì `{((\sqrt{x}-1)^2 ge0),(3\sqrt{x}ge0):}`
`->2/((\sqrt{x}-1)^2+3\sqrt{x})` luôn dương `∀x`
Vậy `Q` luôn dương `∀x (đpcm)`
