Giải thích các bước giải:
\(\begin{array}{l}
a,\frac{{x + 2}}{{2{x^2} - 8}} = \frac{{x + 2}}{{2({x^2} - 4)}} = \frac{{x + 2}}{{2(x - 2)(x + 2)}} = \frac{1}{{2(x - 2)}} = \frac{{x + 3}}{{(x - 2)(x + 3)}}\\
\frac{{4x + 12}}{{{x^2} + 6x + 9}} = \frac{{4(x + 3)}}{{{{(x + 3)}^2}}} = \frac{4}{{(x + 3)}} = \frac{{4(x - 2)}}{{(x - 2)(x + 3)}} = \frac{{4x - 8}}{{(x - 2)(x + 3)}}\\
c,\frac{{x + 3}}{{2{x^2} + 7x + 3}} = \frac{{x + 3}}{{(x + 3)(2x + 1)}} = \frac{1}{{2x + 1}} = \frac{{3(x - 1)}}{{3(x - 1)(2x + 1)}} = \frac{{3x - 3}}{{3(x - 1)(2x + 1)}}\\
\frac{{2x - 2}}{{3{x^2} - 6x + 3}} = \frac{{2(x - 1)}}{{3{{(x - 1)}^2}}} = \frac{2}{{3(x - 1)}} = \frac{{2(2x + 1)}}{{3(x - 1)(2x + 1)}} = \frac{{4x + 2}}{{3(x - 1)(2x + 1)}}\\
\end{array}\)