bt: thực hiện phép tính a, 1/3x-1 – 1/3x+1 – 2x-3/1-9x^2 b, 3x^2+2x+1/X^3-1 – 1-x/x^2+X+1 – 2/x-1 c,3x+1/(x-1)^2 – 1/x+1 + x+3/1-x^2

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1. Đáp án:

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

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