{ log0,25x2 (x+6)/4 <= 1
4–x=t2
(320–t2)(64–t) ≥ 5;
(320–t2)/(64–t) – 5 ≥ 0
(320–t2–5·(64–t))/(64–t) ≥ 0
(–t2+5t)(64–t) ≥ 0
(t·(t–5))/(t–64) ≥ 0
______ [0] ___+____ [5] ___–__ (64) _+__
С учетом t > 0
0 < t ≤ 5 или t > 64
0 < 2–x≤5 или 2–x > 26
–x ≤log25 или –x > 6
x ≥ – log2 5 или x < – 6
(2)
{0,25x2>0 ⇒ x ≠ 0
{0,25x2 ≠ 1 ⇒ x ≠ ± 2
{(x+6)/4>0 ⇒ x > –6
(–6;–2) U(–2;0) U(0;2)U(2;+ ∞ )
Метод рационализации
(0,25x2–1)((x+6)/4–0,25x2) ≤ 0
(x–2)(x+2)(x+6–x2) ≤ 0
(x–2)(x+2)(x+2)(x–3) ≥ 0
__+_ (–2) __+_ (2) ____ (3) _+__
C учетом ОДЗ
(–6;–2)U(–2;0) U(0;2)U(3;+ ∞ )
log25 > log24=2
–log25 < –2
Пересечение решений:
[–log25;–2) U(–2;0) U(0;2)U(3;+ ∞ )