помошите пожалуйста по алгебре решить)))

Ответ
Да запросто. Например, 6.18
log2 x/8 = 15 / [log2 (x/16) - 1]
ОДЗ: x>0 и log2 (x/16) не=1 => x/16 не=2 => x не=32
__ log2 x/8 = log2 x - log2 8 = log2 x - log2 2^3 = log2 x - 3
__ log2 (x/16) = log2 x - log2 16 = log2 x - log2 2^4 = log2 x - 4 =>
__ [log2 (x/16) - 1] = log2 x - 4 - 1 = log2 x - 5
=>
log2 x - 3 = 15 / (log2 x - 5)
(log2 x - 3)(log2 x - 5) = 15
(log2 x)^2 - 3log2 x - 5log2 x + 15 = 15
(log2 x)^2 - 8log2 x = 0
log2 x * (log2 x - = 0
log2 x = 0 => x = 1
log2 x - 8 = 0 => log2 x = 8 => x=2^8 = 256

....
(lg x)^(-1) + 4lg x^2 + 9 = 0
ОДЗ: x>0
1/lg x + 4*2*lg x + 9 = 0
8*(lg x)^2 + 9*lg x + 1 = 0
lg x = t
8t^2 + 9t + 1
t1 = -29 => lg x = -29 - не проходит по ОДЗ.
t2 = 20 => lg x = 20 => x=10^20