Question

Помогите с выполнением заданий 10,11,12. Спасибо.

Answer 1

$10.\; \; \frac{3a-9}{b+2}\cdot \frac{3b+6}{18a-54}=\frac{3(a-3)}{b+2}\cdot \frac{3(b+2)}{18(a-3)}=\frac{1}{2} \\\\11.\; \; \frac{(a-2)^2}{b+5}:\frac{6-3a}{2b+10}=\frac{(a-2)^2}{b+5}\cdot \frac{2(b+5)}{-3(a-2)}=-\frac{2\cdot (a-2)}{3}=\frac{2\cdot (2-a)}{3}\\\\12.\; \; x=-1\\\\ \frac{4}{x^2+4x+4}:(\frac{x^2+12}{x^2-4}-\frac{x+2}{x-2})=\frac{4}{(x+2)^2}:(\frac{x^2+12}{(x-2)(x+2)}-\frac{x+2}{x-2})=\\\\=\frac{4}{(x+2)^2}:\frac{x^2+12-(x+2)^2}{(x+2)(x-2)}=\frac{4}{(x+2)^2}:\frac{x^2+12-x^2-4x-4}{(x+2)(x-2)}=\\\\=\frac{4}{(x+2)^2}\cdot \frac{(x+2)(x-2)}{8-4x}=\frac{4}{x+2}\cdot \frac{x-2}{-4(x-2)}=-\frac{1}{x+2}=-\frac{1}{-1+2}=-1$

Answer 2

$1) \frac{3a-9}{b+2} *\frac{3b+6}{18a-54} =\frac{3(a-3)*3(b+2)}{(bx^{2}2)*18(a-3)}=\frac{9}{18}=\frac{1}{2} \\\\2) \frac{(a-2)^{2}}{b+5} :\frac{6-3a}{2b+10}=\frac{(a-2)^{2}}{b+5} *\frac{2(b+5)}{3(2-a)}=\frac{2(2-a)}{3} \\\\3) \\\\ \frac{x^{2}+12}{x^{2}-4}-\frac{x+2}{x-2}=\frac{x^{2}+12}{(x-2)(x+2)} -\frac{x+2}{x-2}=\frac{x^{2}+12-(x+2)(x+2)}{x^{2}-4}=\frac{x^{2}+12-x^{2}-4x-4}{x^{2}-4}=\frac{8-4x}{x^{2}-4} \\\\\frac{4}{x^{2}+4x+4} :\frac{8-4x}{x^{2}-4}=$$=\frac{4}{(x+2)^{2}} *\frac{(x+2)(x-2)}{8-4x}=-\frac{1}{(x+2)} \\\\\frac{-1}{(-1+2)}=-1$