Cho A= ($\frac{1}{X-1}$ – $\frac{X}{1-X^3}$ . $\frac{X^2+X+1}{X+1}$) : $\frac{2X+1}{X^2+2X+1}$ . a)Rút gọn A. b)Tính A khi x= $\frac{1}{2}$.

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1. Đáp án:

    b) A=-3

   Giải thích các bước giải:

   \(\begin{array}{l}
   a)DK:x \ne  \pm 1\\
   A = \left[ {\dfrac{1}{{x – 1}} – \dfrac{x}{{\left( {1 – x} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 1}}} \right].\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
    = \left[ {\dfrac{1}{{x – 1}} + \dfrac{x}{{\left( {x – 1} \right)\left( {x + 1} \right)}}} \right].\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
    = \dfrac{{x + 1 + x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
    = \dfrac{{2x + 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
    = \dfrac{{x + 1}}{{x – 1}}\\
   b)Thay:x = \dfrac{1}{2}\\
    \to A = \dfrac{{\dfrac{1}{2} + 1}}{{\dfrac{1}{2} – 1}} =  – 3
   \end{array}\)

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