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nelyatargonska
@nelyatargonska
March 2022
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log13(x^2-3)≥log13(x+4)
Очень нужно. Пожалуйста))
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ЕкарныйБабай
Log13(x^2-3)≥log13(x+4)
одз x²-3>0 x<-√3 x>√3
x+4>0 x>-4
x=(-4 -√3)(√3 +00)
x²-3≥x+4
x²-x-7≥0
D=1+28=29
x12=(1+-√29)/2
x≤(1-√29)/2 ≈ -2.19
x≥(1+√29)/2
x=(-4 (1-√29)/2] U [ (1+√29)/2 +00)
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Answers & Comments
одз x²-3>0 x<-√3 x>√3
x+4>0 x>-4
x=(-4 -√3)(√3 +00)
x²-3≥x+4
x²-x-7≥0
D=1+28=29
x12=(1+-√29)/2
x≤(1-√29)/2 ≈ -2.19
x≥(1+√29)/2
x=(-4 (1-√29)/2] U [ (1+√29)/2 +00)