Question

Помогите девятый вопрос ​

Answer 1

Объяснение:

$\frac{3}{b^2-2b+1} -\frac{6}{b^2-1} +\frac{3}{1+2b+b^2}=\frac{3}{(b-1)^2} -\frac{6}{(b+1)*(b-1)} +\frac{3}{b^2+2b+1}=\\= \frac{3}{(b-1)^2} -\frac{6}{(b+1)*(b-1)} +\frac{3}{(b+1)^2}= \frac{3*(b+1)^2-6(b+1)*(b-1)+3*(b-1)^2}{(b-1)^2*(b+1)^2}=\\=\frac{3*(b^2+2b+1)-6*(b^2-1)+3*(b^2-2b+1)}{(b-1)^2*(b+1)^2}=\frac{3b^2+6b+3-6b^2+6+3b^2-6b+3}{(b-1)^2*(b+1)^2} =\\=\frac{12}{(b-1)^2*(b+1)^2}=\frac{12}{((b-1)*(b+1))^2} =\frac{12}{(b^2-1)^2}=\frac{12}{((\sqrt{10})^2-1)^2 }=$

$=\frac{12}{(10-1)^2} =\frac{12}{9^2} =\frac{12}{81}=\frac{4}{27} .$