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CMath
@CMath
March 2022
1
9
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Помогите решить задачу:
Знайти три послідовних цілих числа,якщо подвоєний квадрат першого з них на 26 більший за добуток другого і третього.
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zhenyaM2002
Verified answer
I = n
II = (n + 1)
III = (n + 1 + 1) = (n + 2)
2n² - (n + 1)(n + 2) = 26
2n² - (n² + 2n +n + 2) = 26
2n² - (n² + 3n + 2) = 26
2n² - n² - 3n - 2 = 26
n² - 3n - 2 - 26 = 0
n² - 3n - 28 = 0
D = (-3)² - 4*1*(-28) = 9 + 112 = 121 = 11² ; D>0
n₁ = ( - (-3) - 11)/(2*1) = (3 -11)/2 = -8/2 = - 4 ∉ N
n₂ = (- (-3) + 11)/(2*1) = (3 +11)/2 = 14/2 = 7
7 + 1 = 8
7 + 2 = 9
Ответ : 7, 8, 9.
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Answers & Comments
Verified answer
I = nII = (n + 1)
III = (n + 1 + 1) = (n + 2)
2n² - (n + 1)(n + 2) = 26
2n² - (n² + 2n +n + 2) = 26
2n² - (n² + 3n + 2) = 26
2n² - n² - 3n - 2 = 26
n² - 3n - 2 - 26 = 0
n² - 3n - 28 = 0
D = (-3)² - 4*1*(-28) = 9 + 112 = 121 = 11² ; D>0
n₁ = ( - (-3) - 11)/(2*1) = (3 -11)/2 = -8/2 = - 4 ∉ N
n₂ = (- (-3) + 11)/(2*1) = (3 +11)/2 = 14/2 = 7
7 + 1 = 8
7 + 2 = 9
Ответ : 7, 8, 9.