Task/27003226
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4*5²ˣ +5*4²ˣ =9*20ˣ =0 ;
4*5²ˣ -9*20ˣ + 5*4²ˣ=0 ;
4*( (5/4)ˣ)² -9*(5/4)ˣ +5 =0 ; замена : t =(5/4)ˣ >0
4t² -9t +5 =0 D =9² -4*4*5 =1²
t₁ =(9 -1)/8 =1 ⇔(5/4)ˣ =1 ⇒ x₁ =0 ;
t₂ =(9+1)/8 =5/4⇔5/4)ˣ =5/4 ⇒ x₂ =1.
ответ : 1. * * * 0 +1 =1 * * *
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2⁵ˣ*2¹⁸*3⁴ˣ*3¹¹*7³ˣ*7⁴ ≥2³ˣ*3²ˣ*7ˣ*2²¹*3¹⁴*7⁷ ;
2²ˣ*3²ˣ*7²ˣ ≥ 2³*3³*7³ ;
(2*3*7)²ˣ ≥ (2*3*7)³
2x ≥ 3 ;
x ≥ 1,5.
ответ: x ∈ [1,5 ; +∞) .