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Ответ

Ответ дан
MizoriesKun

log₂₋ₓ (x-2)^2 = 1      ОДЗ  2-х >0  х<2  , 2-х≠ 1   х≠1

(x-2)^2 = 2-х          ⇒ 2-х >0

х²-4х+4 =2-х

х²-3х+2=0

D= 9-8=1

x₁=(3+1)/2= 2 не подходит под ОДЗ

x₂=(3-1)/2= 1 не пододит под ОДЗ

Ответ : х∈∅

log3 (3+x) + log3 (x-5) = 2    ОДЗ 3+х>0  x> -3 , x-5>0 x>5

log3 (3+x)*(x-5) = 2

(3+x)*(x-5) = 3²

3x+x²-15-5x =9

x²-2x-24=0

D=4+96=100

x₁= (2+10)/2=6

x₂=(2-10)/2=-4  не подходит под ОДЗ

Ответ =6

log5 (x+1) - log5 (6x+1) = -1     ОДЗ  х+1>0   х>-1 , 6х+1>0   х > - 1/6

log5 (x+1) / (6x+1) = -1

(x+1) / (6x+1) = 5⁻¹

(x+1) / (6x+1) = 1/5

5*(x+1) =1*(6x+1)

5х+5=6х+1

х=4

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