Если lg5=a, lg7=b, то выразите log 2 по основанию 14 через а и в

Log(a) b = 1 / log(b) a log(a) (b*c) = log(a) b + log(a) c log(a) (b/c) = log(a) b - log(a) c log(a) b = log(c) b / log(c) a log(14) 2 = 1/ log(2) 14 log(2) 14 = log(2) 2 + log(2) 7 = 1 + log(2) 7  log(2) 7 = lg 7 / lg 2 = lg 7 / (lg 10/5) = lg 7 / ( lg 10 - lg 5 ) = lg 7 / ( 1 - lg 5) = b/(1-a) 1 + log(2) 7 = 1 + b/(1-a) = (1+b-a)/(1-a) log(14) 2 = (1-a)/(1+b-a) 

1+2+3+4++(n-1)=156

1+2+3++17=153 команд 18

156 возможно ошибка

1)(6ху+8у)-2ху-8у+1=6ху+8у-2ху-8у+1

2)5с+15-с^[2]-3с-3с+3с^[2]=2c+2c^[2]-6c+15

3)5b^[2]-5bc+2bc-c^[2]=5b^[2]-c^[2]-3bc