решить систему:
3^4х-1 + 3^4х+1 >=80
logx/2 (4x^2 - 3x + 1) >=0
Ответы на вопрос
Ответил yoxset
0
3^(4х-1)+3^(4х+1)>=80
3^(4x)/3^1 + 3^(4x) * 3^1 =>80
3^(4x)=t
t/3 + 3t => 80
t +3*3t => 80*3
10t => 240
t => 24
3^(4x) =>24
log 3(24) <= 4x
x =>1/4 log3(24)
x =>log 3 [24^(1/4)]
3^(4x)/3^1 + 3^(4x) * 3^1 =>80
3^(4x)=t
t/3 + 3t => 80
t +3*3t => 80*3
10t => 240
t => 24
3^(4x) =>24
log 3(24) <= 4x
x =>1/4 log3(24)
x =>log 3 [24^(1/4)]
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