a)x+9/x^2-9-3/x^2+3x b)3/2x+6-x-6/2x^2+6x c)x+1/x-3-1-x/x+3-2x(1-x)/9-x^2

0 bình luận về “a)x+9/x^2-9-3/x^2+3x b)3/2x+6-x-6/2x^2+6x c)x+1/x-3-1-x/x+3-2x(1-x)/9-x^2”

1. Đáp án:

    c) \(\dfrac{2}{{x – 3}}\)

   Giải thích các bước giải:

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

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