Đáp án:
`A={2(\sqrt{x}+1)}/{x+\sqrt{x}+1}` với `x\ge 0;x\ne 1`
Giải thích các bước giải:
`A=({x+2}/{x\sqrt{x}-1}+x/{x+\sqrt{x}+1}+1/{1-\sqrt{x}}):{\sqrt{x}-1}/2`
`\qquad (x\ge 0;x\ne 1)`
`=({x+2}/{(\sqrt{x})^3-1^3}+x/{x+\sqrt{x}+1}-1/{\sqrt{x}-1}). 2/{\sqrt{x}-1}`
`={x+2+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-x-\sqrt{x}-1}/{(\sqrt{x}-1)^2(x+\sqrt{x}+1)} . 2`
`=2.{x\sqrt{x}-x-\sqrt{x}+1}/{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}`
`=2. {x(\sqrt{x}-1)-(\sqrt{x}-1)}/{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}`
`={2(\sqrt{x}-1)(x-1)}/{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}`
`={2(\sqrt{x}-1)(\sqrt{x}-1)(\sqrt{x}+1)}/{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}`
`={2(\sqrt{x}+1)}/{x+\sqrt{x}+1}`
Vậy `A={2(\sqrt{x}+1)}/{x+\sqrt{x}+1}` với `x\ge 0;x\ne 1`
