Question

пж помогите даю 15 баллов​

Answer 1

Ответ:

$\frac{1}{27a + 9}$

Объяснение:

$\frac{a-3}{2a + 4} *\frac{a^{2}-4}{a^{3}-27}*\frac{a^{2}+3a+9}{a^{2}-2a}$ =

$\frac{a-3 * a^{2}-4 * a^{2}+3a+9}{2a + 4 * a^{3}-27 * a^{2}-2a}$ =

$\frac{a^{6} -3 *-4 *+3+9}{2a^{7} + 4 * -27 *-2}$ =

$\frac{-a^{6}+3 * 4 * 12}{2a^{7} + 4 * 27 *2}$ =

$\frac{-4a^{6}+12 * 12}{54a^{7} + 108 *2}$ =

$\frac{-48a^{6}+144}{108a^{7} + 216}$ =

$\frac{-48+144}{108a + 216}$ =

$\frac{144 - 48}{108a + 216}$ =

$\frac{12}{108a + 27}$ =

$\frac{4}{108a + 9}$ =

$\frac{1}{27a + 9}$