44. 1 2c 2 (c-3)²+12c c²+3c+2+c²+4c+3 +c²+5c+6.2

г) Упростим выражение $$\frac{1}{c^2+3c+2}+\frac{2c}{c^2+4c+3} + \frac{2 \cdot (c-3)^2+12c}{2(c^2+5c+6)}$$.

Разложим знаменатели на множители:

$$c^2+3c+2 = (c+1)(c+2)$$.

$$c^2+4c+3 = (c+1)(c+3)$$.

$$c^2+5c+6 = (c+2)(c+3)$$.

Исходное выражение:

$$\frac{1}{(c+1)(c+2)} + \frac{2c}{(c+1)(c+3)} + \frac{2 (c-3)^2+12c}{2(c+2)(c+3)} = \frac{1}{(c+1)(c+2)} + \frac{2c}{(c+1)(c+3)} + \frac{c^2 - 6c + 9+6c}{(c+2)(c+3)} = \frac{1}{(c+1)(c+2)} + \frac{2c}{(c+1)(c+3)} + \frac{c^2 + 9}{(c+2)(c+3)}$$.

Приведем к общему знаменателю:

$$\frac{c+3+2c(c+2) + (c^2+9)(c+1)}{(c+1)(c+2)(c+3)} = \frac{c+3+2c^2+4c + c^3+c^2+9c+9}{(c+1)(c+2)(c+3)} = \frac{c^3 + 3c^2 + 14c + 12}{(c+1)(c+2)(c+3)} = \frac{(c+1)(c^2+2c+12)}{(c+1)(c+2)(c+3)} = \frac{c^2+2c+12}{(c+2)(c+3)}$$.

Ответ: $$\frac{c^2+2c+12}{(c+2)(c+3)}$$