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BlackSoul5
@BlackSoul5
August 2022
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Помогите решить пожалуйста!!!
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kirichekov
Verified answer
3^x=a
a/(a-9)+(a+9)/(a-4)+83/(a-4)(a-9)≤0
(a²-4a+a²-81+83)/(a-9)(a-4)≤0
(2a²-4a+2)/(a-4)(a-9)≤0
2(a²-2a+1)/(a-4)(a-9)≤0
2(a-1)²/(a-4)(a-9)≤0
a=1 a=9 a=4
+ + _ +
----------[1]---------------(4)------------(9)-------------
a=1⇒3^x=1⇒x=0
4<a<9⇒4<3^x<9⇒log(3)4<x<2
x∈(log(3)4;2) U {0}
1 votes
Thanks 1
BlackSoul5
Спасибо,выручил!
kirichekov
пожалуйста, успехов Вам.
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Answers & Comments
Verified answer
3^x=aa/(a-9)+(a+9)/(a-4)+83/(a-4)(a-9)≤0
(a²-4a+a²-81+83)/(a-9)(a-4)≤0
(2a²-4a+2)/(a-4)(a-9)≤0
2(a²-2a+1)/(a-4)(a-9)≤0
2(a-1)²/(a-4)(a-9)≤0
a=1 a=9 a=4
+ + _ +
----------[1]---------------(4)------------(9)-------------
a=1⇒3^x=1⇒x=0
4<a<9⇒4<3^x<9⇒log(3)4<x<2
x∈(log(3)4;2) U {0}