Cộng, trừ phân thức: $\frac{1}{{2x + 2}} - \frac{{x - 1}}{{3{x^2} + 6x + 3}}$.

ĐKXĐ: x ≠ –1

$\frac{1}{{2x + 2}} - \frac{{x - 1}}{{3{x^2} + 6x + 3}} = \frac{1}{{2(x + 1)}} - \frac{{x - 1}}{{3({x^2} + 2x + 1)}}$

$= \frac{1}{{2(x + 1)}} - \frac{{x - 1}}{{3{{(x + 1)}^2}}} = \frac{{3(x + 1)}}{{2(x + 1).3(x + 1)}} - \frac{{2(x - 1)}}{{3{{(x + 1)}^2}.2}}$

$= \frac{{3x + 3}}{{6{{(x + 1)}^2}}} - \frac{{2x - 2}}{{6{{(x + 1)}^2}}} = \frac{{3x + 3 - 2x + 2}}{{6{{(x + 1)}^2}}} = \frac{{x + 5}}{{6{{(x + 1)}^2}}}$.