Đáp án+Giải thích các bước giải:

 Bài `2:`

`a)M=((\sqrt{x})/(\sqrt{x}-1)+(\sqrt{x})/(x-1)):(2/x-(2-x)/(x\sqrt{x}+x))`

`(đk : x\ge0;x\ne1)`

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

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

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

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

`=(x)/(\sqrt{x}-1)`

Vậy `M=(x)/(\sqrt{x}-1)` với `x\ge0;x\ne1`

`b)M=(-1)/(2)`

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

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

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

`\sqrt{x}-1+2x=0`

`2x+2\sqrt{x}-\sqrt{x}-1=0`

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

`(\sqrt{x}+1).(2\sqrt{x}-1)=0`

`\sqrt{x}+1=0` hoặc `2\sqrt{x}-1=0`

`\sqrt{x}=-1` (Vô lí) hoặc `\sqrt{x}=1/2`

`x=(1/2)^{2}`

`x=1/4(tmđk)`

Vậy `x=1/4` để `M=(-1)/(2)`