Задача 26841 ...

u952205146

26 апреля 2018 г.

5 log₂(x – 6 + 6 / (x – 1)) ≤ log√(2)(3 / (x – 4) – 2 / (x – 3)) + 7.

SOVA

29 апреля 2018 г.

ОДЗ:
{(x²–7x+12)/(x–1) > 0

(1;3) U (4;+ ∞)


Так как
log_√2y=1/(1/2)log₂y=2log₂y
y > 0

5log₂(x–6+(6/(x–1))=5log₂(x²–7x+12)/(x–1)


2log₂(3/(x–4)–(2/(x–3))=2log₂(x–1)/(x²–7x+12)


5log₂(x²–7x+12)/(x–1) ≤ 2log₂(x–1)/(x²–7x+12) + 7

5log₂(x–3)+5log₂(x–4)–5log₂(x–1)≤ 2log₂(x–1)–2log₂(x–3)–2log₂(x–3) +7

7log₂(x–4)(x–3)/(x–1) ≤ 7

log₂(x–4)(x–3)/(x–1) ≤ 1

(x–4)(x–3)/(x–1) ≤ 2

(x²–7x+12–2x+2))/(x–1) ≤ 0

(x²–9x+14)/(x–1) ≤ 0

(x–2)(x–7)/(x–1) ≤ 0

_–__ (1) ____ (2) _____–_____ (7) __________

С учетом ОДЗ получаем ответ.
(4;7)