1: \(A=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
2: \(A-6=\dfrac{2x-4\sqrt{x}+2}{\sqrt{x}}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>A>6
`(a):\ A=(2x+2)/(\sqrt{x})+(x\sqrt{x}-1)/(x-\sqrt{x})-(x\sqrt{x}+1)/(x+\sqrt{x})`
`(ĐK:x>0;x\ne 1)`
`=(2x+2)/(\sqrt{x})+((\sqrt{x}-1)(x+\sqrt{x}+1))/(\sqrt{x}(\sqrt{x}-1))-((\sqrt{x}+1)(x-\sqrt{x}+1))/(\sqrt{x}(\sqrt{x}+1))`
`=(2x+2)/(\sqrt{x})+(x+\sqrt{x}+1)/(\sqrt{x})-(x-\sqrt{x}+1)/(\sqrt{x})`
`=(2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1)/(\sqrt{x})`
`=(2x+2\sqrt{x}+2)/(\sqrt{x})`
`(b):\ A-6=(2x+2\sqrt{x}+2)/(\sqrt{x})-6`
`=(2x+2\sqrt{x}+2-6\sqrt{x})/(\sqrt{x})`
`=(2x-4\sqrt{x}+2)/(\sqrt{x})`
`=(2(x-2\sqrt{x}+1))/(\sqrt{x})`
`=(2(\sqrt{x}-1)^{2})/(\sqrt{x})`
Với : `x>0;x\ne 1`
`=>2(\sqrt{x}-1)^{2}>0;\sqrt{x}>0`
Do đó nên : `A-6=(2(\sqrt{x}-1)^{2})/(\sqrt{x})>0`
Hay `A>6`
