`Đáp` `án:`

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

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

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

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

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

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

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

`b) M = -2/(x^2+x+1) = -2/7`

`⇔ -2x^2-2x-2=-14`

`⇔ -2x^2-2x+12=0`

`⇔ -2x^2+4x-6x+12=0`

`⇔ -2x(x-2)-6(x-2)=0`

`⇔ (x-2)(-2x-6)=0`

`⇔` $\left[\begin{matrix} x-2=0\\ -2x-6=0\end{matrix}\right.$

`⇔` $\left[\begin{matrix} x=2 \\ x=-3\end{matrix}\right.$

`Vậy` `S = {2; 3} thì M = -2/7`

`tex{~ Học Tốt Nha ~}`

Đáp án:

a/ Ta có : M=$\frac{3(x²+1)}{x³-1}$ -$\frac{x+1}{x²+x+1}$ - $\frac{2}{x-1}$   

=$\frac{3(x²+1)}{x³-1}$ - $\frac{(x+1)(x-1)}{x³-1}$- $\frac{2(x²+x+1)}{x³-1}$  

=$\frac{3x²+3)}{x³-1}$ -$\frac{x²-1}{x³-1}$- $\frac{2x²+2x+2)}{x³-1}$  

=$\frac{3x²+3-x²+1-2x²-2x-2}{x³-1}$ 

=$\frac{-2x+2}{x³-1}$ 

=$\frac{-(2x-2)}{x³-1}$ 

=$\frac{-2(x-1)}{(x-1)(x²+x+1)}$ 

=$\frac{-2}{x²+x+1}$ 

b/Để M=$\frac{-2}{7}$ 

⇒$\frac{-2}{x²+x+1}$ =$\frac{-2}{7}$ 

⇔ -2(x²+x+1) = -2.7

⇔-2x²-2x-2 = -14

⇔-2x²-2x+12 = 0

⇔-2x²-6x+4x+12=0

⇔-(2x²+6x)+(4x+12)=0

⇔-2x(x+3)+4(x+3)=0

⇔(x+3)(-2x+4)=0

⇒\(\left[ \begin{array}{l}x+3=0\\-2x+4=0\end{array} \right.\) 

⇒\(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)