[text]1)\frac{b_{12}}{b_{8}}=\frac{4}{1}b_{12}=4b_{8}b_{1}*q^{11}=4b_{1}*q^{7} |:b_{1}q^{7}q^{4}=4q=\sqrt[4]{4}=\sqrt{2}
2)(b_{2}-b_{4})^{2}-b_{5}=-\frac{35}{18}(b_{1}q-b_{1}q^{3})^{2}-b_{1}q^{4}=-\frac{35}{18}(b_{1}q)^{2}*(1-q^{2})^{2}-b_{1}q^{4}+\frac{35}{18}=0(b_{1}*\sqrt{2})^{2}*(1-(\sqrt{2})^{2} )^{2}-b_{1}*(\sqrt{2})^{4}+\frac{35}{18}=02b_{1}^{2}-4b_{1}+\frac{35}{18}=036b_{1}^{2}-72b_{1}+35=0D=(-72)^{2}-4*36*35=5184-5040=144=12^{2}b_{1}'=\frac{72-12}{72}=\frac{60}{72}=\frac{5}{6}b_{1}''=\frac{72+12}{72}=\frac{84}{72}=1\frac{1}{6}>1
b_{7}=b_{1}*q^{6}=\frac{5}{6}*(\sqrt{2})^{6}=\frac{5}{6}*8=\frac{20}{3}=6\frac{2}{3}Otvet:\boxed{b_{7}=6\frac{2}{3}}[\text]
а^2-в^2=а-в;
(а+в)(а-в)=а-в;
Разделив правую и левую часть на (а-в), получим: а+в=0
ответ: 0