rút gọn biểu thức a)1-3m/2m + 3m-2/2m-1 + 3m-2/2m-4m^2 c)K/k^2-9 + 1/6k-9-k^2 + 1/k^2+6k+9 g)1/2x-3 – 2/3-2x + 18/9-4x^2 h)1/

0 bình luận về “rút gọn biểu thức a)1-3m/2m + 3m-2/2m-1 + 3m-2/2m-4m^2 c)K/k^2-9 + 1/6k-9-k^2 + 1/k^2+6k+9 g)1/2x-3 – 2/3-2x + 18/9-4x^2 h)1/”

1. Đáp án:

   $\begin{array}{l}
   a)\dfrac{{1 – 3m}}{{2m}} + \dfrac{{3m – 2}}{{2m – 1}} + \dfrac{{3m – 2}}{{2m – 4{m^2}}}\\
    = \dfrac{{1 – 3m}}{{2m}} + \dfrac{{3m – 2}}{{2m – 1}} + \dfrac{{3m – 2}}{{2m\left( {1 – 2m} \right)}}\\
    = \dfrac{{\left( {1 – 3m} \right).\left( {2m – 1} \right) + \left( {3m – 2} \right).2m – 3m + 2}}{{2m\left( {2m – 1} \right)}}\\
    = \dfrac{{2m – 1 – 6{m^2} + 3m + 6{m^2} – 4m – 3m + 2}}{{2m\left( {2m – 1} \right)}}\\
    = \dfrac{{ – 2m + 1}}{{2m\left( {2m – 1} \right)}}\\
    = \dfrac{{ – 1}}{{2m}}\\
   c)\dfrac{k}{{{k^2} – 9}} + \dfrac{1}{{6k – 9 – {k^2}}} + \dfrac{1}{{{k^2} + 6k + 9}}\\
    = \dfrac{k}{{\left( {k – 3} \right)\left( {k + 3} \right)}} – \dfrac{1}{{{k^2} – 6k + 9}} + \dfrac{1}{{{{\left( {k + 3} \right)}^2}}}\\
    = \dfrac{{k\left( {k – 3} \right)\left( {k + 3} \right) – {{\left( {k + 3} \right)}^2} + {{\left( {k – 3} \right)}^2}}}{{{{\left( {k + 3} \right)}^2}{{\left( {k – 3} \right)}^2}}}\\
    = \dfrac{{k\left( {{k^2} – 9} \right) – {k^2} – 6k – 9 + {k^2} – 6k + 9}}{{{{\left( {k + 3} \right)}^2}{{\left( {k – 3} \right)}^2}}}\\
    = \dfrac{{{k^3} – 9k – 12k}}{{{{\left( {k + 3} \right)}^2}{{\left( {k – 3} \right)}^2}}}\\
    = \dfrac{{{k^3} – 21k}}{{{{\left( {k + 3} \right)}^2}{{\left( {k – 3} \right)}^2}}}\\
   g)\dfrac{1}{{2x – 3}} – \dfrac{2}{{3 – 2x}} + \dfrac{{18}}{{9 – 4{x^2}}}\\
    = \dfrac{1}{{2x – 3}} + \dfrac{2}{{2x – 3}} + \dfrac{{18}}{{\left( {3 – 2x} \right)\left( {3 + 2x} \right)}}\\
    = \dfrac{3}{{2x – 3}} – \dfrac{{18}}{{\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\
    = \dfrac{{3\left( {2x + 3} \right) – 18}}{{\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\
    = \dfrac{{6x + 9 – 18}}{{\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\
    = \dfrac{{6x – 9}}{{\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\
    = \dfrac{2}{{2x + 3}}\\
   h)\dfrac{1}{{{x^2} – 5x – 6}} – \dfrac{x}{{x – 6}}\\
    = \dfrac{1}{{\left( {x – 6} \right)\left( {x + 1} \right)}} – \dfrac{x}{{x – 6}}\\
    = \dfrac{{x + 1 – x}}{{\left( {x – 6} \right)\left( {x + 1} \right)}}\\
    = \dfrac{1}{{\left( {x – 6} \right)\left( {x + 1} \right)}}
   \end{array}$

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