Thực hiện phép tính: `a,(4x+1)/2-(3x+2)/3` `b,(x+3)/3-x/(x-3)+9/(x^2-3x)` `c, (x+3)/(x^2-1)-1/(x^2+x)`

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1. Đáp án:

    d) \(\dfrac{{x + 1}}{{x\left( {x – 1} \right)}}\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   a)\dfrac{{3\left( {4x + 1} \right) – 2\left( {3x + 2} \right)}}{6}\\
    = \dfrac{{12x + 3 – 6x – 4}}{6}\\
    = \dfrac{{6x – 1}}{6}\\
   b)\dfrac{{\left( {x + 3} \right)\left( {{x^2} – 3x} \right) – 3{x^2} + 27}}{{3x\left( {x – 3} \right)}}\\
    = \dfrac{{{x^3} – 3{x^2} + 3{x^2} – 9x – 3{x^2} + 27}}{{3x\left( {x – 3} \right)}}\\
    = \dfrac{{{x^3} – 3{x^2} – 9x + 27}}{{3x\left( {x – 3} \right)}}\\
    = \dfrac{{{x^2}\left( {x – 3} \right) – 9\left( {x – 3} \right)}}{{3x\left( {x – 3} \right)}}\\
    = \dfrac{{{{\left( {x – 3} \right)}^2}\left( {x + 3} \right)}}{{3x\left( {x – 3} \right)}} = \dfrac{{{x^2} – 9}}{{3x}}\\
   c)\dfrac{{x\left( {x + 3} \right) – \left( {x – 1} \right)}}{{x\left( {x + 1} \right)\left( {x – 1} \right)}}\\
    = \dfrac{{{x^2} + 3x – x + 1}}{{x\left( {x + 1} \right)\left( {x – 1} \right)}}\\
    = \dfrac{{{x^2} + 2x + 1}}{{x\left( {x + 1} \right)\left( {x – 1} \right)}}\\
    = \dfrac{{{{\left( {x + 1} \right)}^2}}}{{x\left( {x + 1} \right)\left( {x – 1} \right)}} = \dfrac{{x + 1}}{{x\left( {x – 1} \right)}}
   \end{array}\)

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