Đáp án:
1) \(\dfrac{{6\left( {x + 2} \right)}}{{x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
1)\dfrac{{3\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}.\dfrac{{2\left( {{x^2} - 4} \right)}}{{x + 2}}\\
= \dfrac{{3\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}.\dfrac{{2\left( {x - 2} \right)\left( {x + 2} \right)}}{{x + 2}}\\
= \dfrac{{6\left( {x + 2} \right)}}{{x - 2}}\\
2)\dfrac{{{x^2} + 2x + x + 2}}{{{x^2}\left( {x + 2} \right) - \left( {x + 2} \right)}} = \dfrac{{x\left( {x + 2} \right) + \left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {{x^2} - 1} \right)}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{1}{{x - 1}}\\
3)\dfrac{{x\left( {x + y} \right)}}{{5\left( {{x^2} - {y^2}} \right)}}.\dfrac{{3\left( {{x^3} - {y^3}} \right)}}{{x\left( {x - y} \right)}}\\
= \dfrac{{x\left( {x + y} \right)}}{{5\left( {x - y} \right)\left( {x + y} \right)}}.\dfrac{{3\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}{{x\left( {x - y} \right)}}\\
= \dfrac{{3\left( {{x^2} + xy + {y^2}} \right)}}{{5\left( {x - y} \right)}}
\end{array}\)
