Question

Помогите пожалуйста ​

Answer 1

$\frac{64b^{2}-128b+64 }{b} :(\frac{4}{b} -4)=\frac{4(16b^{2}-32b+16) }{b}:\frac{4-4b}{b}=\frac{4(4b-4)^{2}}{b}*\frac{b}{4-4b}=\\\\=4(4-4b)=16(1-b)\\\\b=-\frac{15}{16}\\\\16(1-b)=16*(1+\frac{15}{16} )=16*\frac{31}{16} =31\\\\Otvet:\boxed{31}$