tính biểu thức a (4x/x^2-4+2x-4/x+2)*x+2/2x+2/2-x

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1. Đáp án:$A = \frac{{x – 2}}{x}$

    

   Giải thích các bước giải: $ĐKXĐ:\{ _{x \ne  \pm 2}^{x \ne 0}$

   $\begin{array}{l}
   A = \left( {\frac{{4x}}{{{x^2} – 4}} + \frac{{2x – 4}}{{x + 2}}} \right).\frac{{x + 2}}{{2x}} + \frac{2}{{2 – x}}\\
    = \left[ {\frac{{4x + (2x – 4)(x – 2)}}{{(x – 2)(x + 2)}}} \right].\frac{{x + 2}}{{2x}} – \frac{2}{{x – 2}}\\
    = \frac{{4x + 2{x^2} – 8x + 8}}{{(x + 2)(x – 2)}}.\frac{{x + 2}}{{2x}} – \frac{2}{{x – 2}}\\
    = \frac{{2{x^2} – 4x + 8}}{{2x(x – 2)}} – \frac{2}{{x – 2}}\\
    = \frac{{{x^2} – 2x + 4}}{{x(x – 2)}} – \frac{2}{{x – 2}}\\
    = \frac{{{x^2} – 2x + 4 – 2x}}{{x(x – 2)}}\\
    = \frac{{{x^2} – 4x + 4}}{{x(x – 2)}}\\
    = \frac{{{{(x – 2)}^2}}}{{x(x – 2)}}\\
    = \frac{{x – 2}}{x}
   \end{array}$

    

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