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vasilisa2107
@vasilisa2107
July 2022
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Прошу помощи! СРОЧНО, ПОЖАЛУЙСТА!
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sedinalana
Решаем 1 неравенство
log(3)(3^x +4)*log(9)(3^(x+1)+12)≥3
log(3)(3^x+4)*log(3)[3(3^x+4)]/log(3)9≥3
log(3)(3^x+4)*(log(3)3+log(3)(3^x+4))/2≥3
log(3)(3^x+4)*(1+log(3)(3^x+4))≥6
log(3)(3^x+4)=a
a(1+a)-6≥0
a+a²-6≥0
D=1+24=25>0
a1+a2=-1 U a1*a2=-6
a1=-3 U a2=2
+ _ +
-------------[-3]-----------------[2]---------------
a≤-3⇒log(3)(3^x+4)≤-3⇒3^x+4≤1/27⇒3^x≤-3 26/27 нет решения
a≥2⇒log(3)(3^x+4)≥2⇒3^x+4≥9⇒3^x≥5⇒x≥log(3)5
Решаем 2 неравенство
5^x²-2*5^(x²-1)-3*5^(x²-2)≤60
5^(x²-2)*(5²-2*5-3)≤60
5^(x²-2)*12≤60
5^(x²-2)≤5
x²-2≤1
x²-3≤0
(x-√3)(x+√3)≤0
+ _ +
-------------[-√3]----------------[√3]----------------
-√3≤x≤√3
Общее
///////////////////////////////////////////////////////
--------[-√3]--------------[log(3)5]--------------[√3]-------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[log(3)5;√3]
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Thanks 1
vasilisa2107
Благодарю!
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Answers & Comments
log(3)(3^x +4)*log(9)(3^(x+1)+12)≥3
log(3)(3^x+4)*log(3)[3(3^x+4)]/log(3)9≥3
log(3)(3^x+4)*(log(3)3+log(3)(3^x+4))/2≥3
log(3)(3^x+4)*(1+log(3)(3^x+4))≥6
log(3)(3^x+4)=a
a(1+a)-6≥0
a+a²-6≥0
D=1+24=25>0
a1+a2=-1 U a1*a2=-6
a1=-3 U a2=2
+ _ +
-------------[-3]-----------------[2]---------------
a≤-3⇒log(3)(3^x+4)≤-3⇒3^x+4≤1/27⇒3^x≤-3 26/27 нет решения
a≥2⇒log(3)(3^x+4)≥2⇒3^x+4≥9⇒3^x≥5⇒x≥log(3)5
Решаем 2 неравенство
5^x²-2*5^(x²-1)-3*5^(x²-2)≤60
5^(x²-2)*(5²-2*5-3)≤60
5^(x²-2)*12≤60
5^(x²-2)≤5
x²-2≤1
x²-3≤0
(x-√3)(x+√3)≤0
+ _ +
-------------[-√3]----------------[√3]----------------
-√3≤x≤√3
Общее
///////////////////////////////////////////////////////
--------[-√3]--------------[log(3)5]--------------[√3]-------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[log(3)5;√3]