Решите срочно!!!! 4^x-5*2^x+1+16≤0. Created by giyosfariza. algebra-ru

Решите срочно!!!! 4^x-5*2^x+1+16≤0...

giyosfariza @giyosfariza

July 2022 1 5 Report

Решите срочно!!!!
4^x-5*2^x+1+16≤0

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DariosI

Verified answer

$2 ^{2x} -5*2*2 ^{x} +16 \leq 0$
$2 ^{2x} -10*2 ^{x} +16 \leq 0$
Пусть $2 ^{x}$ =y
y²-10y+16≤0
D=100-4*16=36=6²
y₁=(10-6)/2=2
y₂=(10+6)/2=8
(y-2)(y-8)≤0
y∈[2; 8]
Значит
$2 ^{x}$ ∈[2; 8]
x∈[1; 3]

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giyosfariza July 2022 | 0 Ответы

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