Đáp án+Giải thích các bước giải:

$(*)$

 `A=1+(\frac{2x+\sqrt{x}-1}{1-x}-\frac{2x\sqrt{x}-\sqrt{x}+x}{1-x\sqrt{x}}).\frac{x-\sqrt{x}}{2\sqrt{x}-1}`

`(đk: x\ge0;x\ne 1;x\ne 1/4)`

`=1+[\frac{-(2x+2\sqrt{x}-\sqrt{x}-1)}{(\sqrt{x})^{2}-1^{2}}+\frac{2x\sqrt{x}-\sqrt{x}+x}{(\sqrt{x})^{3}-1^{3}}].\frac{x-\sqrt{x}}{2\sqrt{x}-1}`

`=1+[\frac{-[2\sqrt{x}.(\sqrt{x}+1)-1.(\sqrt{x}+1)]}{(\sqrt{x}-1).(\sqrt{x}+1)}+\frac{2x\sqrt{x}-\sqrt{x}+x}{(\sqrt{x}-1).(x+\sqrt{x}+1)}].\frac{x-\sqrt{x}}{2\sqrt{x}-1}`

`=1+[\frac{-(2\sqrt{x}-1).(\sqrt{x}+1)}{(\sqrt{x}-1).(\sqrt{x}+1)}+\frac{2x\sqrt{x}-\sqrt{x}+x}{(\sqrt{x}-1).(x+\sqrt{x}+1)}].\frac{x-\sqrt{x}}{2\sqrt{x}-1}`

`=1+[\frac{1-2\sqrt{x}}{\sqrt{x}-1}+\frac{2x\sqrt{x}-\sqrt{x}+x}{(\sqrt{x}-1).(x+\sqrt{x}+1)}].\frac{x-\sqrt{x}}{2\sqrt{x}-1}`

`=1+[\frac{(1-2\sqrt{x}).(x+\sqrt{x}+1)+2x\sqrt{x}-\sqrt{x}+x}{(\sqrt{x}-1).(x+\sqrt{x}+1)}].\frac{x-\sqrt{x}}{2\sqrt{x}-1}`

`=1+[\frac{x+\sqrt{x}+1-2x\sqrt{x}-2x-2\sqrt{x}+2x\sqrt{x}-\sqrt{x}+x}{(\sqrt{x}-1).(x+\sqrt{x}+1)}].\frac{x-\sqrt{x}}{2\sqrt{x}-1}`

`=1+\frac{(-2\sqrt{x}+1).(x-\sqrt{x})}{(\sqrt{x}-1).(x+\sqrt{x}+1).(2\sqrt{x}-1)}`

`=1+\frac{-(2\sqrt{x}-1).\sqrt{x}.(\sqrt{x}-1)}{(\sqrt{x}-1).(x+\sqrt{x}+1).(2\sqrt{x}-1)}`

`=1+\frac{-\sqrt{x}}{x+\sqrt{x}+1}`

`=\frac{x+\sqrt{x}+1-\sqrt{x}}{x+\sqrt{x}+1}`

`=\frac{x+1}{x+\sqrt{x}+1}`

Vậy `A=\frac{x+1}{x+\sqrt{x}+1}` với `x\ge0;x\ne1;x\ne1/4.`

$(*)$

`A=\frac{6-\sqrt{6}}{5}`

`\frac{x+1}{x+\sqrt{x}+1}=\frac{6-\sqrt{6}}{5}` `(đk:x\ge0;x\ne1;x\ne 1/4)`

`(6-\sqrt{6}).(x+\sqrt{x}+1)=5.(x+1)`

`6x+6\sqrt{x}+6-x\sqrt{6}-\sqrt{6x}-\sqrt{6}-5x-5=0`

`x+6\sqrt{x}+1-x\sqrt{6}-\sqrt{6x}-\sqrt{6}=0`

`(x-x\sqrt{6})+(1-\sqrt{6})+(6\sqrt{x}-\sqrt{6x})=0`

`x.(1-\sqrt{6})+1.(1-\sqrt{6})-\sqrt{6x}.(1-\sqrt{6})=0`

`(1-\sqrt{6}).(x-\sqrt{6x}+1)=0`

`x-\sqrt{6x}+1=0`

`(\sqrt{x})^{2}-2.\sqrt{x}. \frac{\sqrt{6}}{2}+(\frac{\sqrt{6}}{2})^{2}-1/2=0`

`(\sqrt{x}-\frac{\sqrt{6}}{2})^{2}=1/2`

`(\sqrt{x}-\frac{\sqrt{6}}{2})^{2}=(\pm\frac{\sqrt{2}}{2})^{2}`

`+)TH1:\sqrt{x}-\frac{\sqrt{6}}{2}=\frac{\sqrt{2}}{2}`

`\sqrt{x}=\frac{\sqrt{2}+\sqrt{6}}{2}`

`\sqrt{x}=\frac{\sqrt{2}.(1+\sqrt{3})}{(\sqrt{2})^{2}}`

`\sqrt{x}=\frac{1+\sqrt{3}}{\sqrt{2}}`

`x=\frac{(1+\sqrt{3})^{2}}{2}`

`x=\frac{4+2\sqrt{3}}{2}`

`x=2+\sqrt{3}(tmđk)`

`+)TH2: \sqrt{x}-\frac{\sqrt{6}}{2}=-\frac{\sqrt{2}}{2}`

`\sqrt{x}=\frac{\sqrt{6}-\sqrt{2}}{2}`

`\sqrt{x}=\frac{\sqrt{2}.(\sqrt{3}-1)}{(\sqrt{2})^{2}}`

`\sqrt{x}=\frac{\sqrt{3}-1}{\sqrt{2}}`

`x=\frac{(\sqrt{3}-1)^{2}}{2}`

`x=\frac{4-2\sqrt{3}}{2}`

`x=2-\sqrt{3}(tmđk)`

Vậy `x\in{2+\sqrt{3}:2-\sqrt{3}}` để `A=\frac{6-\sqrt{6}}{5}.`