[tex] (\frac{1}{3})^{x^2-4x-1} \geq 9^{x-1} [/tex]. Created by guvirmaxim1. algebra-ru

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July 2022 2 10 Report

[tex] (\frac{1}{3})^{x^2-4x-1} \geq 9^{x-1} [/tex]

reshator2000 (1/3)^(x²-4x-1)≥9^(x-1)
3^(-(x²-4x-1)≥3^(2*(x-1)
3^(-x²+4x+1)≥3^(2x-2)
-x²+4x+1≥2x-2
x²-2x-3≤0
x²-2x-3=0 D=16
x₁=3 x₂=-1
(x-3)(x-1)≤0
-∞_____+_____-1_____-______3______+_____+∞
x∈[-1;3].
Ответ: x∈[-1;3].

sangers1959

Verified answer

-x^2+4x+1>x-1
-x^2+3x+2>0
x^2-3x-2<0
d=17
x1=-0.5
x2=3.5
по методу интервалов
x∈[-0.5.3.5}

reshator2000 да я увидел ошибку

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