B= (x^2+ 2/x^3-1 + x/x^2+x+1 + 1/1-x) : x-1/2) a)tìm đk xác định b) RG c)CM B>0

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1. 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:

   $x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=(x+\dfrac{1}{2})^2+\dfrac{3}{4}≥\dfrac{3}{4}>0$

   $⇒\dfrac{2}{x^2+x+1}>0(dpcm)$

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2. Giải thích các bước giải:

   \(\begin{array}{l}
   B = (\frac{{{x^2} + 2}}{{{x^3} – 1}} + \frac{x}{{{x^2} + x + 1}} + \frac{1}{{1 – x}}):\frac{{x – 1}}{2}\\
   a,DKXD:x \ne 1\\
   b,B = (\frac{{{x^2} + 2}}{{{x^3} – 1}} + \frac{{x(x – 1)}}{{{x^3} – 1}} – \frac{{{x^2} + x + 1}}{{{x^3} – 1}}).\frac{2}{{x – 1}}\\
   B = \frac{{{x^2} + 2 + {x^2} – x – {x^2} – x – 1}}{{{x^3} – 1}}.\frac{2}{{x – 1}}\\
   B = \frac{{{x^2} – 2x + 1}}{{{x^3} – 1}}.\frac{2}{{x – 1}}\\
   B = \frac{{{{(x – 1)}^2}}}{{(x – 1)({x^2} + x + 1)}}.\frac{2}{{x – 1}}\\
   B = \frac{2}{{{x^2} + x + 1}}\\
   c,{x^2} + x + 1 = {(x + \frac{1}{2})^2} + \frac{3}{4} > 0\forall x\\
    \Rightarrow B > 0\forall x
   \end{array}\)

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