Question

Решение 6 задания,даю 20 баллов

Answer 1

$\frac{64b^2+128b+64}{b}:(\frac{4}{b}+b)=\frac{(8b)^2+2b+8^2)}{b}:\frac{4+b^2}{b}=$

$\frac{(8b+8)^2}{b}\cdot\frac{b}{4+b}=\frac{(8b+8)^2}{4+b^2}$

$b=-\frac{15}{16}$

$\frac{(8\cdot \left(-\frac{15}{16} \right) +8)^2}{4+ \left( -\frac{15}{16}\right)^2}=\frac{ \left(-\frac{15}{2} +8\right) ^2}{4+ \frac{225}{256}}=$

$\frac{ \left(-\frac{15}{2} +\frac{16}{2}\right) ^2}{4\frac{225}{256}}=\frac{ \left(\frac{1}{2}\right) ^2}{\frac{1249}{256}}=\frac{1}{4}\cdot\frac{256}{1249}$