(log5 2+ log2 5+2)*(log5 2-lg2)*log2 5-log5 2

(log5(2)+log2(5)+2)*(log5(2)-lg2)*log2(5)-log5(2)=

1ая скобка (log5(2)+log2(5)+2)=(log5(2))^2+1+2log5(2)) / log5(2)

2ая скобка (log5(2)-lg2)=1 / log2(5)-1 / (log2(5)+1)=(log2(5)+1-log2(5)) / (log2(5)*(log2(5)+1)=1 / (log2(5) * (log2(5)+1)

3 действие log2(5) * 1 / (log2(5)*(log2(5)+1)=1 / (log2(5)+1) = log5(2) / (1+log5(2))

4 действие (log5(2))^2+1+2log5(2)) / log5(2) * log5(2) / (1+log5(2))=

(log5(2))^2+1+2log5(2)) - квардрат суммы

= (log5(2)+1)^2 / (log5(2)+1) = log5(2)+1

log5(2)+1-log5(2) = 1

ответ: 1.

sn=b1*(q^n-1)/(q-1)=2/81*728/2=728/81

х1- один корень,х2 - второй,х2=6х1. применим теорему виета

х1+6х1=-2

х1*6х1=q, решить систему

7х1=-2

х1=-2/7

-2/7*6*(-2/7)=q

q=24/49