решите плиз №2в ,№6 №7

1) 36^[log 6(5) + log 9 (81)] =
= 36^[log 6(5) + log 9(9^2)] =
= 36^[log 6(5) + 2 * log 9(9) =
= (6^2)^[log 6(5) + 2] =
= (6^2)^[log 6(5)] * (6^2)^2 =
= 6^[2*log 6(5)] * 6^4 =
= 6^[log 6(5^2)] * 6^4 =
= 5^2 * 6^4 = 25 * 1296 = 32400

2)
a) lg x - lg 12 = log 0,1(x+1) - log 100(4)
ОДЗ: (x не = 0) и (х не = -1)
--- log 0,1(x+1) =
= log (1/100)(x+1) =
= log {10^(-2)}(x+1) = 1/(-2) * lg (x+1) = lg (x+1)^(-1/2)
--- log 100(4) = log(10^2)(4) = 1/2 lg (4)= lg 4^(1/2) = lg (2)
lg x - lg 12 = lg (x+1)^(-1/2) - lg 2
x / 12 = (x+1)^(-1/2) / 2
x / 12 = 1/2*(x+1)^(1/2)
2x * (x+1) = 12
2x^2 + 2x - 12 = 0
x^2 + x - 12 = 0
x(1) = -4
x(2) = 3

б) log^2 3(x-1) - 2 log(1/3)[9/(x-1)] = 2^[log 2(7)]
ОДЗ: х не = 1
--- log(1/3)[9/(x-1)] =
= log [3^(-1)] [9/(x-1)] =
=[1/(-1)] log 3 [9/(x-1)] =
= (-1) log 3 [9/(x-1)] =
= log 3 [9/(x-1)]^(-1) = log 3 [(x-1)/9] =
= log 3(x-1) - log 3(9) = log 3(x-1) - log 3(3^2) =
= log 3(x-1) - 2
--- 2^[log 2(7)] = 7
log^2 3(x-1) - 2 * [log 3(x-1) - 2] = 7
log^2 3(x-1) - 2 log 3(x-1) +4 = 7
log^2 3(x-1) - 2 log 3(x-1) - 3 = 0
log 3(x+1) = t
t^2 - 2t - 3 = 0
t(1) = 3
t(2) = -1
log 3(x+1) = t(1)
log 3(x+1) = 3 => 3^3 = x+1 => x=1-9= -8
log 3(x+1) = t(2)
log 3(x+1) = -1 => 3^(-1) = x+1 => x=1-1/3=2/3

6) log (5+x)(1-2x) > log (5+x)3 + log(5+x)x^2
ОДЗ: х не = 0; 5; 1/2
1 - 2x > 3*x^2
3x^2 + 2x - 1 < 0
3x^2 + 2x - 1
x(1) = -1
x(2) = 1/3
(x+1)(x-1/3) <0
-- x+1<0 => x< -1
-- x+1/2 >0 => x> -1/2
- (бесконечность < -1 и -1/2<x<(+бесконечность) >0 => x> -1
-- x+1/2 <0 => x< -1/2
- 1 < x < -1/2

не правельно записан оответ

господи, кто решал выше, прошу вас, научитесь правильно выявлять и записывать область допустимых значений, смотреть противно