(2/3)^(x^2+4x)>=(8/27)^(x+2)

Question

(2/3)^(x^2+4x)>=(8/27)^(x+2)

Answer 1

( 2/3) ^ ( X^2 + 4X) ≥ (2/3) ^ ( 3 * ( X + 2 )) 
X^2 + 4X ≥ 3X + 6 
X^2 + 4X - 3X - 6 ≥ 0 
X^2 + X - 6 ≥ 0 
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D = 1 + 24 = 25 ; V D = 5 
X1 = ( - 1 + 5 ) : 2 = 2 
X2 = ( - 6 ) : 2 = ( - 3 )
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X1 ≥ 2 
X2 ≥ - 3 
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Ответ  Х ∈ [ - 3 ; 2 ]