Задача 14300 ...

u46166761

13.03.2017

log 49 (x+4) + log (x^2 + 8x +16) √7 < = -3/4

sumets6152

14.03.2017

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ОДЗ:
{x2–3 > 0;
{x+9 > 0;
{x2–6x+9 > 0;
{log2(x2–3) ≥ 0;⇒ x2–3 ≥ 1
{log2(x+9) ≥ 0; ⇒ x+9 ≥ 1
{log2(x2–6x+9)≠1; ⇒ x2–6x+9≠1
Из системы
{x2–4≥ 0
{x+8 ≥ 0
{x2–6x+9≠0
{x2–6x+8≠0
получаем
ОДЗ:х∈[–8;–2]U[2;3)U(3;4)U(4;+ ∞).

Дробь ≥0 в двух случаях
1)
{√log2(x2–3) – √log2(x+9)≥0
{log2(x2–6x+9) > 0
2)
{√log2(x2–3) – √log2(x+9)≤0
{log2(x2–6x+9) < 0

или
1)
{√log2(x2–3)≥ √log2(x+9)
{log2(x2–6x+9) > log21
2)
{√log2(x2–3)≤√log2(x+9)
{log2(x2–6x+9) < log21

1)
{log2(x2–3)≥ log2(x+9)
{x2–6x+9 > 1
2)
{log2(x2–3) ≤log2(x+9)
{x2–6x+9 < 1

1)
{x2–3≥ x+9
{x2–6x+9 > 1
2)
{x2–3 ≤x+9
{x2–6x+9 < 1

1)
{x2–x–12≥0
{x2–6x+8 > 0
2)
{x2–x–12 ≤0
{x2–6x+8 < 0

1)
{(–∞;–3]U[4;+∞)
{(–∞;2)U(4;+∞)
C учетом ОДЗ получаем [–8; –3]U(4;+∞)

2)
{[–3;4]
{(2;4)
C учетом ОДЗ получаем (2;3)U(3;4)
О т в е т. [–8; –3]U(2;3)U(3;4)U(4;+∞)