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isit98
@isit98
July 2022
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Помогите решить неравенство
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MizoriesKun
㏒₁₆(х+5)+ ㏒₍ₓ²₊₁₀ₓ₊₂₅₎2 ≥ 3/4
ОДЗ х+5 >0
x > -5
x²+10x+25 > 0 и x²+10x+25≠1
(х+5)²> 0 при любом значении х
x²+10x+25= 1
x²+10x+24= 0
D=100-96=4
x₁=(-10+2)/2 =-4
x₂=(-10-2)/2=-6
ОДЗ :
x∉(-5;-4)∪(-4;+∞)
㏒(₂⁴)(х+5) + 1/㏒₂(х+5)² ≥ 3/4
(1/4) ㏒₂(х+5) + 1/(2㏒₂(х+5) ) - 3/4 ≥ 0 домножим на 4
㏒₂(х+5) + 2/(㏒₂(х+5) ) - 3 ≥ 0 теперь домножим на ㏒₂(х+5)
㏒²₂(х+5) - 3㏒₂(х+5) +2 ≥ 0 замена ㏒₂(х+5) =а
а² -3а+2 ≥ 0
D= 9-8=1
a₁= (3+1)/2=2 ㏒₂(х+5) =2 x+5=4 x₁=-1
a₂=(3-1)/2 =1 ㏒₂(х+5) = 1 x+5=2 x₂=-3
+ - +
- 5______-3________ -1__________
х∈
(-5;-4)∪(-4;-3]∪[-1;+∞)
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Answers & Comments
ОДЗ х+5 >0 x > -5
x²+10x+25 > 0 и x²+10x+25≠1
(х+5)²> 0 при любом значении х
x²+10x+25= 1
x²+10x+24= 0
D=100-96=4
x₁=(-10+2)/2 =-4
x₂=(-10-2)/2=-6
ОДЗ : x∉(-5;-4)∪(-4;+∞)
㏒(₂⁴)(х+5) + 1/㏒₂(х+5)² ≥ 3/4
(1/4) ㏒₂(х+5) + 1/(2㏒₂(х+5) ) - 3/4 ≥ 0 домножим на 4
㏒₂(х+5) + 2/(㏒₂(х+5) ) - 3 ≥ 0 теперь домножим на ㏒₂(х+5)
㏒²₂(х+5) - 3㏒₂(х+5) +2 ≥ 0 замена ㏒₂(х+5) =а
а² -3а+2 ≥ 0
D= 9-8=1
a₁= (3+1)/2=2 ㏒₂(х+5) =2 x+5=4 x₁=-1
a₂=(3-1)/2 =1 ㏒₂(х+5) = 1 x+5=2 x₂=-3
+ - +
- 5______-3________ -1__________
х∈(-5;-4)∪(-4;-3]∪[-1;+∞)