C=$\frac{5x+1}{x³-1}$ - $\frac{1-2x}{x²+x+1}$ - $\frac{2}{1-x}$ (x $\neq$ 1)
=$\frac{5x+1}{(x-1)(x²+x+1)}$ - $\frac{1-2x}{x²+x+1}$ - $\frac{2}{1-x}$
=$\frac{5x+1 - (1-2x)(x-1) + 2(x²+x+1)}{(x-1)(x²+x+1)}$
=$\frac{5x+1-x+2x+1-2x+2x²+2x+2}{(x-1)(x²+x+1)}$
=$\frac{2x²+6x+4}{(x-1)(x²+x+1)}$
=$\frac{2(x²+3x+2)}{(x-1)(x²+x+1)}$
=$\frac{2(x²+x+2x+2)}{(x-1)(x²+x+1)}$
=$\frac{2[x(x+1)+2(x+1)]}{(x-1)(x²+x+1)}
=$\frac{2(x+1)(x+2)}{(x-1)(x²+x+1)}$
b.C>0
⇔$\frac{2(x+1)(x+2)}{(x-1)(x²+x+1)}$ > 0
⇔(x-1)(x²+x+1)>0(vì 2(x+1)(x+2)>0)
⇔x-1>0(vì x²+x+1>0)
⇔x>1
Vậy x>1 thì C>0
Mình chỉ biết làm như vậy thoi xin 5s ạ
