$\textit{Đáp án + Giải thích các bước giải:}$

`a) C=(1/(x-1)-x/(1-x^3).(x^2+x+1)/(x+1)):(2x+1)/(x^2+2x+1)` `(x \ne +-1)`

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

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

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

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

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

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

 $b)$ Với `x=1/2` thì

$C=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-1}$

$=\dfrac{\dfrac{3}{2}}{\dfrac{-1}{2}}$

$=-3$

Vậy với `x=1/2` thì `C=-3`

  ĐKXĐ : ` x ` $\ne$ ` ±1 `

a) ` C = ( ` $\dfrac{1}{ x - 1 }$ - $\dfrac{ x }{1 - x³}$ . $\dfrac{x² + x + 1}{ x + 1 }$) : $\dfrac{2x + 1}{x² + 2x + 1}$

` C = ` ( $\dfrac{1}{ x - 1 }$ + $\dfrac{ x }{x³ - 1}$ . $\dfrac{x² + x + 1}{ x + 1 }$ ) : $\dfrac{2x + 1}{x² + 2x + 1}$

` C= ` ( $\dfrac{1}{ x - 1 }$ + $\dfrac{ x }{ ( x - 1 ) ( x² + x + 1 )}$ . $\dfrac{x² + x + 1}{ x + 1 }$) : $\dfrac{2x + 1}{x² + 2x + 1}$

` C = ` ( $\dfrac{1}{ x - 1 }$ + $\dfrac{ x }{x - 1}$ . $\dfrac{1}{ x + 1 }$) : $\dfrac{2x + 1}{x² + 2x + 1}$

` C = ` ( $\dfrac{1}{ x - 1 }$ + $\dfrac{ x }{( x - 1 )( x + 1 )}$ ) : $\dfrac{2x + 1}{x² + 2x + 1}$

` C = ` ( $\dfrac{x + 1}{ ( x + 1 ) ( x - 1 ) }$ + $\dfrac{ x }{( x - 1 )( x + 1 )}$ ) : $\dfrac{2x + 1}{x² + 2x + 1}$

` C = `  $\dfrac{ 2x + 1}{ ( x - 1 )( x + 1 ) }$  : $\dfrac{2x + 1}{x² + 2x + 1}$

` C = ` $\dfrac{ 2x + 1}{ ( x - 1 ) ( x + 1 ) }$  . $\dfrac{ x² + 2x + 1 }{ 2x + 1 }$

` C = ` $\dfrac{x² + 2x + 1}{ ( x - 1 ) ( x + 1 ) }$

` C = ` $\dfrac{ ( x + 1 )² }{ ( x - 1 ) ( x + 1 ) }$

` C = ` $\dfrac{ x + 1 }{ x - 1 }$

b) Thay `x =` $\dfrac{1}{2}$ vào ` C ` , ta được : 

   $\dfrac{ \dfrac{1}{2} + 1 }{ \dfrac{1}{2} - 1 }$

= $\dfrac{3}{2}$ : ( $\dfrac{-1}{2}$ ) 

= ` -3 `

 Vậy ` C = -3 ` khi ` x = ` $\dfrac{1}{2}$