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Somanda
@Somanda
July 2022
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1)найти модуль суммы корней уравнения:
lg(x^2+9)+lg2=1+lg(x+1)
2)Вычислить:
log8 log4 log2 (8^4*4^2)
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MizoriesKun
lg(x^2+9)+lg2=1+lg(x+1)
ОДЗ
x+1>0 х>-1
lg2*(x^2+9)=lg10*(x+1)
2*(x^2+9)=10*(x+1)
2x²+18-10x-10=0
2x²-10x+8=0 |2
x²-5x+4=0
D=25-16=9 √D=3
x₁=(5+3)/2=4
x₂=(5-3)/2=1
|∑x₁,x₂|=|4+1|=5
log₈ log₄ log₂ (8⁴*4²)=log₈ log₄ log₂ ((2³)⁴*(2²)²)=log₈ log₄ log₂2¹²⁺⁴=
=log₈ log₄ log₂2¹⁶=log₈ log₄ 16=log₈ log₄ 4²=log₈ 2 =log₂³2 = (1/3)*log₂2 =1/3
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MizoriesKun
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MizoriesKun
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Answers & Comments
ОДЗ x+1>0 х>-1
lg2*(x^2+9)=lg10*(x+1)
2*(x^2+9)=10*(x+1)
2x²+18-10x-10=0
2x²-10x+8=0 |2
x²-5x+4=0
D=25-16=9 √D=3
x₁=(5+3)/2=4
x₂=(5-3)/2=1
|∑x₁,x₂|=|4+1|=5
log₈ log₄ log₂ (8⁴*4²)=log₈ log₄ log₂ ((2³)⁴*(2²)²)=log₈ log₄ log₂2¹²⁺⁴=
=log₈ log₄ log₂2¹⁶=log₈ log₄ 16=log₈ log₄ 4²=log₈ 2 =log₂³2 = (1/3)*log₂2 =1/3