Задача 33021 [block](2^(4x) - 2^(3x+1) + 2^(2x+1) -...

sova

25 января 2019 г. в 19:45

Замена переменной
2^x=t
t > 0
2^x+1=2^x·2=2t;
2^2x+1=2^2x·2=2t²
2^3x+1=2^3x·2=2t²

(t⁴ –2t³+2t²–2t+1)/((t–2)²+(t–3)³–1) ≥ 0
(t–1)·(t³–t²+t–1)/((t–2)²+(t–3)³–1) ≥ 0
(t–1)·(t–1)·(t²+1)/((t–2)²+(t–3)³–1) ≥ 0
метод интервалов.
Нули числителя t₁=t₂=1
Нули знаменателя:
(t–2)²+(t–3)³–1=0
t²–4t+4+t³–9t²+27t–27–1=0
t³–8t²+23t–24=0
t³–27 –(8t²–24t)–(t–3)=0
(t–3)·(t²–5t+8)=0
t=3
D=25–4·8<0

(0) _–__ [1] __–___ (3) __+___

t=1 или t > 3
2^x=2 или 2^x > 3
x=1 или x > log₂3

О т в е т. {1}U(log₂3;+ ∞ )