Question

Умоляю решите буду очень рада

Answer 1

$2z^{2}-3z+1=0\\\\D=(-3)^{2} -4*2*1=9-8=1\\\\z_{1}=\frac{3+1}{4}=1\\\\z_{2}=\frac{3-1}{4}=\frac{1}{2}\\\\\\(b+\frac{a^{2}}{b}):(a+2b)= \frac{b^{2}+a^{2}}{b}*\frac{1}{a+2b}=\frac{b^{2}+a^{2}}{b(a+2b)} \\\\1)z=\frac{b}{a}\\\\z_{1}=1\Rightarrow \frac{b}{a}=1\Rightarrow a=b\\\\\frac{b^{2}+a^{2}}{b(a+2b)}= \frac{b^{2}+b^{2}}{b(b+2b)}=\frac{2b^{2}}{3b^{2}}=\frac{2}{3}\\\\2)z_{2}=\frac{1}{2} \Rightarrow\frac{b}{a}=\frac{1}{2}\Rightarrow a=2b$

$\frac{b^{2}+a^{2}}{b(a+2b)}=\frac{b^{2}+(2b)^{2}}{b(2b+2b)}=\frac{b^{2}+4b^{2}}{b*4b}=\frac{5b^{2}}{4b^{2}} =\frac{5}{4}=1,25\\\\1,25\frac{2}{3}\\\\Otvet:\boxed{1,25}$