a, Để B xác định
$⇔x\neq1$
b, $B=(\dfrac{x^2+2}{x^3-1}+\dfrac{x}{x^2+x+1}+\dfrac{1}{1-x}):\dfrac{x-1}{2}$ $(x\neq1)$
$B=(\dfrac{x^2+2}{(x-1)(x^2+x+1)}+\dfrac{x}{x^2+x+1}-\dfrac{1}{x-1})\cdot\dfrac{2}{x-1}$
$B=\dfrac{x^2+2+x(x-1)-x^2-x-1}{(x-1)(x^2+x+1)}\cdot\dfrac{2}{x-1}$
$B=\dfrac{x^2+2+x^2-x-x^2-x-1}{(x-1)(x^2+x+1)}\cdot\dfrac{2}{x-1}$
$B=\dfrac{x^2-2x+1}{(x-1)(x^2+x+1)}\cdot\dfrac{2}{x-1}$
$B=\dfrac{2(x-1)^2}{(x-1)^2(x^2+x+1)}$
$B=\dfrac{2}{x^2+x+1}$
Vậy $B=\dfrac{2}{x^2+x+1}$ với $x\neq1$
c, Ta có:
$x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=(x+\dfrac{1}{2})^2+\dfrac{3}{4}≥\dfrac{3}{4}>0$
$⇒\dfrac{2}{x^2+x+1}>0(dpcm)$
