Задача 37081 ...

vk221889286

15 мая 2019 г. в 17:08

Решите неравенство
2^x + 2^x+2/(2^x –4) + (4^x+ 7·2^x+20)/(4^x –3·2^x+2 +32) ≤ 1

sova

15 мая 2019 г. в 17:42

2^x=t

t+4t/(t–4) + (t²+7t+20)/(t²–12t+32) ≤ 1

t + 4t/(t–4) + (t²+7t+20)/((t–4)(t–8)) – 1 ≤ 0

(t·(t²–12t+32)+4t·(t–8)+(t²+7t+20)–t²+12t–32 )/((t–4)(t–8)) ≤ 0

(t³–8t²+19t–12)/((t–4)(t–8)) ≤ 0

t=1 – корень многочлена t³–8t²+19t–12, значит

t³–8t²+19t–12=(t–1)(t²–7t+12)=(t–1)·(t–3)(t–4)


(t–1)·(t–3)(t–4)/((t–4)(t–8)) ≤ 0

(t–1)·(t–3)/(t–8) ≤ 0
t ≠ 4

__–__ [1] __+__ [3] _– _ (4) ____–_____ (8) ___+__

t ≤ 1 или 3 ≤ t < 4 или 4 < t < 8

2^x ≤ 1 или 3 ≤2^x< 4 или 4 <2^x< 8

2^x ≤ 2⁰ или 2^log₂3 ≤2^x<2² или 2² <2^x< 2³

О т в е т. (– ∞;0] U[ log₂3;2) U (2;3)