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July 2022
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nKrynka
Решение
√(x⁴ + 2x - 5) = 1 + x
(√(x⁴ + 2x - 5)² = (1 + x)²
1 + x ≥ 0, x ≥ - 1, x ∈ [- 1; + ∞)
x⁴ + 2x - 5 = 1 + 2x + x²
x⁴ - x² - 5 - 1 = 0
x⁴ - x² - 6 = 0
x² = t
t² - t - 6 = 0
t₁ = 3
t₂ = - 2
x² = 3
x₁ = - √3 не удовлетворяет условию x ∈ [- 1; + ∞)
x₂ = √3
Ответ: √3
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Answers & Comments
√(x⁴ + 2x - 5) = 1 + x
(√(x⁴ + 2x - 5)² = (1 + x)²
1 + x ≥ 0, x ≥ - 1, x ∈ [- 1; + ∞)
x⁴ + 2x - 5 = 1 + 2x + x²
x⁴ - x² - 5 - 1 = 0
x⁴ - x² - 6 = 0
x² = t
t² - t - 6 = 0
t₁ = 3
t₂ = - 2
x² = 3
x₁ = - √3 не удовлетворяет условию x ∈ [- 1; + ∞)
x₂ = √3
Ответ: √3