Question

Задание 3 с решением, СРОЧНО

Answer 1

$\frac{a+4}{a-1}:\Big (\frac{9a-9}{3a+4}-\frac{(2a-7)^2}{3a^2+a-4}\Big )-\frac{2}{a-2}=\frac{a+4}{a-1}:\Big (\frac{9(a-1)}{3a+4}-\frac{(2a-7)^2}{(3a+4)(a-1)}\Big )-\frac{2}{a-2}=\\\\\\=\frac{a+4}{a-1}:\frac{9(a-1)^2-(2a-7)^2}{(3a+4)(a-1)}-\frac{2}{a-2}=\frac{a+4}{a-1}:\frac{(3(a-1)-(2a-7))\cdot (3(a-1)+2a-7)}{(3a+4)(a-1)}-\frac{2}{a-2}=\\\\\\=\frac{a+4}{a-1}:\frac{(a+4)\cdot (5a-10)}{(3a+4)(a-1)}-\frac{2}{a-2}=\frac{a+4}{a-1}\cdot \frac{(3a+4)(a-1)}{5\, (a+4)(a-2)}-\frac{2}{a-2}=$

$=\frac{3a+4}{5\, (a-2)}-\frac{2}{a-2}=\frac{3a+4-10}{5\, (a-2)}=\frac{3a-6}{5\, (a-2)}=\frac{3\, (a-2)}{5\, (a-2)}=\frac{3}{5}$