Log4 (x + 1) + log4 x log4 2 - id2052458520211222 от Polina19051 22.12.2021 00:07

Question

Математика: Log4 (x + 1) + log4 x log4 2

Polina19051 · Accepted Answer

(29 - х): 3 - 1 = 6
(29 - х): 3 = 6 + 1
(29 - х) : 3 = 7
29 - х = 7 * 3
29 - х = 21
х = 29 - 21
х = 8

(29 - 8) : 3 - 1 = 6
21 : 3 - 1 =6
7 - 1 = 6
6 = 6

-4х - 8(9 - 2х)= - 4х - 8*9 - 8 *(-2х) = - 4х -72 + 16х =
= (16х - 4х) - 72 = 12х -72 =
можно еще вынести общий множитель:
= 12(х - 6)

6(3х-2)+4(5х-1) = 6*3х + 6*(-2) + 4*5х + 4*(-1) =
= 18х - 12 + 20х - 4 = (18х +20х) - (12 + 4) =
= 38х - 16 =
= 2(19х - 8)

-6(8а-2b) +(3a-7b)*2 = -6*8a - 6*(-2b) + 3a*2 -7b * 2 =
= -48a + 12b + 6a - 14b = - (48a - 6a) - (14b - 12b) =
= - 42a -2b =
= - 2(21a +b)

4(8t+7) - 16(2t-3) = 4*8t + 4*7 - 16*2t - 16*(-3) =
= 32t + 28 - 32t + 48 = 28 + 48 = 76

Log4 (x + 1) + log4 x < log4 2

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