trừ các phân thức sau 1/3x-2-4/3x+2-3x-6/4×9x^2

0 bình luận về “trừ các phân thức sau 1/3x-2-4/3x+2-3x-6/4×9x^2”

1. Đáp án:

   $\begin{array}{l}
   \frac{1}{{3x – 2}} – \frac{4}{{3x + 2}} – \frac{{3x – 6}}{{4 – 9{x^2}}}\\
    = \frac{1}{{3x – 2}} – \frac{4}{{3x + 2}} + \frac{{3x – 6}}{{\left( {3x – 2} \right)\left( {3x + 2} \right)}}\\
    = \frac{{3x + 2 – 4\left( {3x – 2} \right) + 3x – 6}}{{\left( {3x – 2} \right)\left( {3x + 2} \right)}}\\
    = \frac{{3x + 2 – 12x + 8 + 3x – 6}}{{\left( {3x – 2} \right)\left( {3x + 2} \right)}}\\
    = \frac{{4 – 6x}}{{\left( {3x – 2} \right)\left( {3x + 2} \right)}}\\
    = \frac{{ – 2\left( {3x – 2} \right)}}{{\left( {3x – 2} \right)\left( {3x + 2} \right)}}\\
    = \frac{{ – 2}}{{3x + 2}}
   \end{array}$

   Bình luận