`a)B=1/x-x/(2x+1)+(2x^2-3x-1)/(x(2x+1))(x\ne0,x\ne-1/2)` 

`=(2x+1)/(x(2x+1))-x^2/(x(2x+1))+(2x^2-3x-1)/(x(2x+1))`

`=(2x+1-x^2+2x^2-3x-1)/(x(2x+1))`

`=(x^2-x)/(x(2x+1))/(x(2x+1))`

`=(x(x-1))/(x(2x+1))`

`=(x-1)/(2x+1)`

`b)C=A:B`

`=(x-1)/x^2:(x-1)/(2x+1)(x\ne0,x\ne-1/2;x\ne1)`

`=(x-1)/x^2*(2x+1)/(x-1)`

`=(2x+1)/x^2`

`C>=-1` suy ra: `(2x+1)/x^2>=-1`

`(2x+1)/x^2+1>=0`

`(x^2+2x+1)/x^2>=0`

`(x+1)^2/x^2>=0`

Ta có: `x^2>0\AAx\ne0` 

Và: `(x+1)^2>=0\AAx`

Suy ra: `(x+1)^2/x^2>=0\AAx`

Vậy `C>=-1` khi `x\ne0,x\ne-1/2,x\ne1`