Đáp án:
a) \(\dfrac{3}{{x - 3}} = \dfrac{{3x}}{{x\left( {x - 3} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{2}{{x\left( {x - 3} \right)}}\\
\dfrac{3}{{x - 3}} = \dfrac{{3x}}{{x\left( {x - 3} \right)}}\\
b)\dfrac{{{x^2} + x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
\dfrac{2}{{{x^2} + x + 1}} = \dfrac{{2\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{2x - 2}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
\dfrac{1}{{x - 1}} = \dfrac{{{x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
c)\dfrac{4}{{x + 2}} = \dfrac{{4\left( {x - 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} = \dfrac{{4x - 8}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
\dfrac{2}{{x - 2}} = \dfrac{{2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} = \dfrac{{2x + 4}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
\dfrac{{5x - 6}}{{4 - {x^2}}} = - \dfrac{{5x - 6}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
d)\dfrac{3}{{2x}} = \dfrac{{3\left( {2x - 1} \right)}}{{2x\left( {2x - 1} \right)}} = \dfrac{{6x - 3}}{{2x\left( {2x - 1} \right)}}\\
\dfrac{{3x - 2}}{{2x - 1}} = \dfrac{{2x\left( {3x - 2} \right)}}{{2x\left( {2x - 1} \right)}} = \dfrac{{6{x^2} - 4x}}{{2x\left( {2x - 1} \right)}}\\
\dfrac{{3x - 2}}{{2x - 4{x^2}}} = - \dfrac{{3x - 2}}{{2x\left( {2x - 1} \right)}}
\end{array}\)
