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Arthurchke
@Arthurchke
August 2022
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24/(x^2-2x)=12/(x^2-x)+x^2-x
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Dимасuk
Verified answer
ОДЗ:
x ≠ 0; 1; 2
24/(x² - 2x) = 12/(x² - x) + x² - x
24/(x² - 2x) - 12/(x² - x) = x² - x
(24x² - 24x - 12x² + 24x)/(x(x - 1)(x - 2) = x² - x
12x²/x²(x - 1)(x - 2) = x(x - 1)
12/(x - 1)(x - 2) = x(x - 1)
12 = x(x - 2)(x - 1)²
12 = (x² - 2x)(x² - 2x + 1)
Пусть t = x² - 2x.
12 = t(t + 1)
t² + t - 12 = 0
t₁ + t₂ = -1
t₁t₂ = -12
t₁ = -4; t₂ = 3
Обратная замена:
x² - 2x = 3
x² - 2x - 3 = 0
x² - 2x + 1 - 4 = 0
(x - 1)² - 2² = 0
(x - 1 - 2)(x - 1 + 2) = 0
(x - 3)(x + 1) = 0
x = -1; 3
Вторая замена:
x² - 2x = -4
x² - 2x + 1 + 4 - 1 = 0
(x - 1)² = -3 - нет корней, т.к. квадрат всегда неотрицательный
Ответ: x = -1; 3.
2 votes
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Answers & Comments
Verified answer
ОДЗ:x ≠ 0; 1; 2
24/(x² - 2x) = 12/(x² - x) + x² - x
24/(x² - 2x) - 12/(x² - x) = x² - x
(24x² - 24x - 12x² + 24x)/(x(x - 1)(x - 2) = x² - x
12x²/x²(x - 1)(x - 2) = x(x - 1)
12/(x - 1)(x - 2) = x(x - 1)
12 = x(x - 2)(x - 1)²
12 = (x² - 2x)(x² - 2x + 1)
Пусть t = x² - 2x.
12 = t(t + 1)
t² + t - 12 = 0
t₁ + t₂ = -1
t₁t₂ = -12
t₁ = -4; t₂ = 3
Обратная замена:
x² - 2x = 3
x² - 2x - 3 = 0
x² - 2x + 1 - 4 = 0
(x - 1)² - 2² = 0
(x - 1 - 2)(x - 1 + 2) = 0
(x - 3)(x + 1) = 0
x = -1; 3
Вторая замена:
x² - 2x = -4
x² - 2x + 1 + 4 - 1 = 0
(x - 1)² = -3 - нет корней, т.к. квадрат всегда неотрицательный
Ответ: x = -1; 3.