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July 2022
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СРОЧНО!
сумма абсцисс критических точек функции f(x)=x^3+12x 2+21x-6 равна...?
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Беня2018
F'(x)=3x²+24x+21=0
3x²+24x+21=0 разделим на 3
х²+8х+7=0
в=64-28=36
x1=(-8+6)/2=-1
x2=(-8-6)/2=-7
сумма абсцисс -7-1=-8
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Answers & Comments
3x²+24x+21=0 разделим на 3
х²+8х+7=0
в=64-28=36
x1=(-8+6)/2=-1
x2=(-8-6)/2=-7
сумма абсцисс -7-1=-8