A=3x - 2
B = x - 2
√A > B аналогично совокупности систем
1. A ≥ 0
B < 0
2. B≥ 0
A > B²
√(3x - 2) > x - 2
решаем 1
x - 2 < 0
3x - 2 ≥ 0
x < 2
x ≥ 2/3
x ∈ [2/3,2)
решаем 2
x - 2 ≥ 0 x ≥ 2
3x - 2 > (x - 2)²
3x - 2 > x² - 4x + 4
x² - 7x + 6 < 0
D = 49 - 24 = 25
x12 = (7 +- 5)/2 = 1 6
(x - 1)(x - 6) > 0
+++++++(1) ------------ (6) ++++++++
x ∈ [2,6)
объединяем 1 и 2
x ∈ [2/3, 6)
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Verified answer
A=3x - 2
B = x - 2
√A > B аналогично совокупности систем
1. A ≥ 0
B < 0
2. B≥ 0
A > B²
√(3x - 2) > x - 2
решаем 1
x - 2 < 0
3x - 2 ≥ 0
x < 2
x ≥ 2/3
x ∈ [2/3,2)
решаем 2
x - 2 ≥ 0 x ≥ 2
3x - 2 > (x - 2)²
3x - 2 > x² - 4x + 4
x² - 7x + 6 < 0
D = 49 - 24 = 25
x12 = (7 +- 5)/2 = 1 6
(x - 1)(x - 6) > 0
+++++++(1) ------------ (6) ++++++++
x ∈ [2,6)
объединяем 1 и 2
x ∈ [2/3, 6)