Đáp án:
e) \(\dfrac{1}{{{x^2} + 6x + 9}} = \dfrac{{x + 5}}{{{{\left( {x + 3} \right)}^2}\left( {x + 5} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
b)\dfrac{5}{{2\left( {x - 2} \right)}} = \dfrac{{5.15}}{{30\left( {x - 2} \right)}} = \dfrac{{75}}{{30\left( {x - 2} \right)}}\\
\dfrac{4}{{3\left( {x - 2} \right)}} = \dfrac{{4.10}}{{30\left( {x - 2} \right)}} = \dfrac{{40}}{{30\left( {x - 2} \right)}}\\
\dfrac{7}{{5\left( {2 - x} \right)}} = - \dfrac{{7.6}}{{30\left( {x - 2} \right)}} = - \dfrac{{42}}{{30\left( {x - 2} \right)}}\\
c)\dfrac{3}{{2\left( {x + 2} \right)}} = \dfrac{{3\left( {x - 2} \right)}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}} = \dfrac{{3x - 6}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
\dfrac{{5x}}{{2\left( {2 - x} \right)}} = - \dfrac{{5x\left( {x + 2} \right)}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}} = - \dfrac{{5{x^2} + 10x}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\dfrac{{10}}{{\left( {2 - x} \right)\left( {x + 2} \right)}} = - \dfrac{{10.2}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}} = - \dfrac{{20}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}\\
d)\dfrac{1}{{{x^3} + 1}} = \dfrac{1}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
\dfrac{2}{{x + 1}} = \dfrac{{2\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \dfrac{{2{x^2} - 2x + 2}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
\dfrac{3}{{{x^2} - x + 1}} = \dfrac{{3\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \dfrac{{3x + 3}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
e)\dfrac{1}{{{x^2} + 8x + 15}} = \dfrac{1}{{\left( {x + 3} \right)\left( {x + 5} \right)}} = \dfrac{{x + 3}}{{{{\left( {x + 3} \right)}^2}\left( {x + 5} \right)}}\\
\dfrac{1}{{{x^2} + 6x + 9}} = \dfrac{1}{{{{\left( {x + 3} \right)}^2}}} = \dfrac{{x + 5}}{{{{\left( {x + 3} \right)}^2}\left( {x + 5} \right)}}
\end{array}\)
( câu a đề bạn xem lại nhé )
