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dashasolodovni1
@dashasolodovni1
July 2022
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Решите,пожалуйста,неравенство)
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hraky
(1 2/7)^(x²-4)≤1
x²-4≤0
(x-2)(x+2)≤0
x=2 x=-2
+ _ +
------------[-2]------------[2]--------------
x∈[-2;2]
1 votes
Thanks 1
sedinalana
Verified answer
(9/7)^(x²-4)<=(9/7)^0
(x²-4)<=0
(х-2)(х+2)<=0
x₁=-2
x₂=2
методом интервалов х ∈ [-2;2]
0 votes
Thanks 0
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Answers & Comments
x²-4≤0
(x-2)(x+2)≤0
x=2 x=-2
+ _ +
------------[-2]------------[2]--------------
x∈[-2;2]
Verified answer
(9/7)^(x²-4)<=(9/7)^0(x²-4)<=0
(х-2)(х+2)<=0
x₁=-2
x₂=2
методом интервалов х ∈ [-2;2]