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deevil
@deevil
July 2022
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Помогите решить!!!Срочно!!!!!!!! Только В
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sedinalana
4*3^x-9*2^x-5*3^(x/2)*2^(x/2)≥0/2^x
4*(3/2)^x-5*(3/2)^(x/2)-9≥0
(3/2)^(x/2)=a
4a²-5a-9≥0
D=25+144=169
a1=(5-13)/8=-1 U a2=(5+13)/8=9/4
+ _ +
------------------[-1]------------------[9/4]--------------------
a≤1⇒(3/2)^(x/2)≤1⇒x/2≤0⇒x≤0
a≥9/4⇒(3/2)^(x/2)≥9/4⇒x/2≥2⇒x≥4
x∈(-∞;0] U [4;∞)
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Answers & Comments
4*(3/2)^x-5*(3/2)^(x/2)-9≥0
(3/2)^(x/2)=a
4a²-5a-9≥0
D=25+144=169
a1=(5-13)/8=-1 U a2=(5+13)/8=9/4
+ _ +
------------------[-1]------------------[9/4]--------------------
a≤1⇒(3/2)^(x/2)≤1⇒x/2≤0⇒x≤0
a≥9/4⇒(3/2)^(x/2)≥9/4⇒x/2≥2⇒x≥4
x∈(-∞;0] U [4;∞)