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irademidovad
@irademidovad
July 2022
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Решите уравнение
(x^2-x+1)-10(x-4)(x+3)-109=0
В ответ запишите сумму корней.
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(x²-x+1)-10(x-4)(x+3)-109=0
(x²-x+1)-10(x²-4x+3x-12)-109=0
x²-x+1-10(x²-x-12)-109=0
x²-x+1-10x²+10x+120-109=0
-9x²+9x+12=0
3x²-3x-4=0, D=9+48=57, √D=√57
x1=(3+√57)/6
x2=(3-√57)/6
x1+x2= (3+√57)/6+(3-√57)/6=(3+√57+3-√57)/6=6/6=1
Otvet: 1
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Answers & Comments
(x²-x+1)-10(x²-4x+3x-12)-109=0
x²-x+1-10(x²-x-12)-109=0
x²-x+1-10x²+10x+120-109=0
-9x²+9x+12=0
3x²-3x-4=0, D=9+48=57, √D=√57
x1=(3+√57)/6
x2=(3-√57)/6
x1+x2= (3+√57)/6+(3-√57)/6=(3+√57+3-√57)/6=6/6=1
Otvet: 1