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laharley
@laharley
June 2022
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Решите уравнение
(х+3)^3-(х+1)^3=56
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vsem2privet
Раскладываем как разность кубов, получаем
(х+3-х-1)(3х^2+4x+13)=56
2(3x^2+4x+13)=56
3x^2+4x+13=28
3x^2+4x-15=0
D=4-(-15*3)=49
x=(-2+-7)/6=5/6; -3/2
ОТВЕТ: 5/6 и -3/2.
0 votes
Thanks 1
egeniyegorov1
(a+b)^3=a^3+3a^2*b+3*a*b^2+b^3
x^3+9x^2+27x+27-x^3-3x^2-3x-1=56
6x^2+24x-30=0
x^2+4x-5=0
D=36
x1=1 x2=-5
1 votes
Thanks 1
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Answers & Comments
(х+3-х-1)(3х^2+4x+13)=56
2(3x^2+4x+13)=56
3x^2+4x+13=28
3x^2+4x-15=0
D=4-(-15*3)=49
x=(-2+-7)/6=5/6; -3/2
ОТВЕТ: 5/6 и -3/2.
x^3+9x^2+27x+27-x^3-3x^2-3x-1=56
6x^2+24x-30=0
x^2+4x-5=0
D=36
x1=1 x2=-5