3x+1/(x-1)^2 – 1/x+1 + x+3/1-x^2

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1. Giải thích các bước giải:

   \(\begin{array}{l}
   \dfrac{{3x + 1}}{{{{\left( {x – 1} \right)}^2}}} – \dfrac{1}{{x + 1}} + \dfrac{{x + 3}}{{1 – {x^2}}}\\
    = \dfrac{{3x + 1}}{{{{\left( {x – 1} \right)}^2}}} – \dfrac{1}{{\left( {x + 1} \right)}} – \dfrac{{\left( {x + 3} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
    = \dfrac{{3x\left( {x + 1} \right) – {{\left( {x – 1} \right)}^2} – \left( {x + 3} \right)\left( {x – 1} \right)}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
    = \dfrac{{3{x^2} + 3x – {x^2} + 2x – 1 – {x^2} – 2x + 3}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
    = \dfrac{{{x^2} + 3x + 2}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}} = \dfrac{{{x^2} + 2x + x + 2}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
    = \dfrac{{x\left( {x + 2} \right) + \left( {x + 2} \right)}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}} = \dfrac{{\left( {x + 2} \right)\left( {x + 1} \right)}}{{{{\left( {x – 1} \right)}^2}\left( {x + 1} \right)}}\\
    = \dfrac{{x + 2}}{{{{\left( {x – 1} \right)}^2}}}
   \end{array}\)

    

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