$\frac{16}{(a-2{)}^4}:(\frac{1}{(a-2{)}^2}-\frac{2}{(a^2-4)}+\frac{1}{(a+2{)}^2})-\frac{8a}{(a-2{)}^2}=1$
$\frac{16}{(a-2{)}^4}:(\frac{1}{(a-2{)}^2}-\frac{2}{(a-2)(a+2)}+\frac{1}{(a+2{)}^2})-\frac{8a}{(a-2{)}^2}=1$
$\frac{16}{(a-2{)}^4}:\frac{(a+2{)}^2-2(a-2)(a+2)+(a-2{)}^2}{(a-2{)}^2(a+2{)}^2}-\frac{8a}{(a-2{)}^2}=1$
$\frac{16}{(a-2{)}^4}:\frac{(a+2-(a-2){)}^2}{(a-2{)}^2(a+2{)}^2}-\frac{8a}{(a-2{)}^2}=1$
$\frac{16}{(a-2{)}^4}:\frac{(a+2-a+2{)}^2}{(a-2{)}^2(a+2{)}^2}-\frac{8a}{(a-2{)}^2}=1$
$\frac{16}{(a-2{)}^4}:\frac{4^2}{(a-2{)}^2(a+2{)}^2}-\frac{8a}{(a-2{)}^2}=1$
$\frac{16}{(a-2{)}^4}\ast \frac{(a-2{)}^2(a+2{)}^2}{16}-\frac{8a}{(a-2{)}^2}=1$
$\frac{1}{(a-2{)}^2}\ast \frac{(a+2{)}^2}{1}-\frac{8a}{(a-2{)}^2}=1$
$\frac{(a+2{)}^2-8a}{(a-2{)}^2}=1$
$\frac{a^2+4a+4-8a}{(a-2{)}^2}=1$
$\frac{a^2-4a+4}{(a-2{)}^2}=1$
$\frac{(a-2{)}^2}{(a-2{)}^2}=1$
1 = 1

$\frac{a+11}{a+9}-(\frac{a+5}{a^2-81}+\frac{a+7}{a^2-18a+81}):(\frac{a+3}{a-9}{)}^2=1$
$\frac{a+11}{a+9}-(\frac{a+5}{(a-9)(a+9)}+\frac{a+7}{(a-9{)}^2}):\frac{(a+3{)}^2}{(a-9{)}^2}=1$
$\frac{a+11}{a+9}-\frac{(a+5)(a-9)+(a+7)(a+9)}{(a-9{)}^2(a+9)}\ast \frac{(a-9{)}^2}{(a+3{)}^2}=1$
$\frac{a+11}{a+9}-\frac{a^2+5a-9a-45+a^2+7a+9a+63}{a+9}\ast \frac{1}{(a+3{)}^2}=1$
$\frac{a+11}{a+9}-\frac{2a^2+12a+18}{a+9}\ast \frac{1}{(a+3{)}^2}=1$
$\frac{a+11}{a+9}-\frac{2(a^2+6a+9)}{a+9}\ast \frac{1}{(a+3{)}^2}=1$
$\frac{a+11}{a+9}-\frac{2(a+3{)}^2}{a+9}\ast \frac{1}{(a+3{)}^2}=1$
$\frac{a+11}{a+9}-\frac{2}{a+9}=1$
$\frac{a+11-2}{a+9}=1$
$\frac{a+9}{a+9}=1$
1 = 1