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/x^2-5x-6 - x/x-6

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}$