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ANTYnoob
@ANTYnoob
July 2022
1
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Найдите сумму всех действительных корней уравнения (1+x+x^2)(1+x+…+x^10)=(1+x+…+x6)^2
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nelle987
Verified answer
1 + x + x^2 = (x^3 - 1)/(x - 1)
1 + x + x^2 + ... + x^10 = (x^11 - 1)/(x - 1)
1 + x + x^2 + ... + x^6 = (x^7 - 1)/(x - 1)
(x^3 - 1)(x^11 - 1) / (x - 1)^2 = (x^7 - 1)^2 / (x - 1)^2
(x^3 - 1)(x^11 - 1) = (x^7 - 1)^2
x^14 - x^11 - x^3 + 1 = x^14 - 2x^7 + 1
x^11 - 2x^7 + x^3 = 0
x^3 * (x^8 - 2x^4 + 1) = 0
x^3 * (x^4 - 1)^2 = 0
x^3 = 0 или x^4 - 1 = 0
x = 0 или x = +-1
Проверка:
x = 0: 1 * 1 = 1^2 - верно
x = 1: 3 * 11 = 7^2 - неверно
x = -1: 1 * 1 = 1^2 - верно
Сумма корней 0 + (-1) = -1.
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Answers & Comments
Verified answer
1 + x + x^2 = (x^3 - 1)/(x - 1)1 + x + x^2 + ... + x^10 = (x^11 - 1)/(x - 1)
1 + x + x^2 + ... + x^6 = (x^7 - 1)/(x - 1)
(x^3 - 1)(x^11 - 1) / (x - 1)^2 = (x^7 - 1)^2 / (x - 1)^2
(x^3 - 1)(x^11 - 1) = (x^7 - 1)^2
x^14 - x^11 - x^3 + 1 = x^14 - 2x^7 + 1
x^11 - 2x^7 + x^3 = 0
x^3 * (x^8 - 2x^4 + 1) = 0
x^3 * (x^4 - 1)^2 = 0
x^3 = 0 или x^4 - 1 = 0
x = 0 или x = +-1
Проверка:
x = 0: 1 * 1 = 1^2 - верно
x = 1: 3 * 11 = 7^2 - неверно
x = -1: 1 * 1 = 1^2 - верно
Сумма корней 0 + (-1) = -1.