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crmss
@crmss
July 2022
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Математика, 11 класс
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IrkaShevko
Verified answer
√(x-1) = t, t ≥ 0
|t - 2| + |t - 3| = 1
1) 0 ≤ t ≤ 2
2 - t + 3 - t = 1
5 - 2t = 1
-2t = -4
t = 2
√(x-1) = 2
x-1 = 4
x = 5
2) 2 < x < 3
t - 2 + 3 - t = 1
1 = 1 - верно
2 < √(x-1) < 3
4 < x-1 < 9
5 < x < 10 - бесконечное число решений
3) x ≥ 3
t - 2 + t - 3 = 1
2t - 5 = 1
2t = 6
t = 3
√(x-1) = 3
x-1 = 9
x = 10
Ответ: x∈[5;10] - бесконечное число решений (Д)
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Answers & Comments
Verified answer
√(x-1) = t, t ≥ 0|t - 2| + |t - 3| = 1
1) 0 ≤ t ≤ 2
2 - t + 3 - t = 1
5 - 2t = 1
-2t = -4
t = 2
√(x-1) = 2
x-1 = 4
x = 5
2) 2 < x < 3
t - 2 + 3 - t = 1
1 = 1 - верно
2 < √(x-1) < 3
4 < x-1 < 9
5 < x < 10 - бесконечное число решений
3) x ≥ 3
t - 2 + t - 3 = 1
2t - 5 = 1
2t = 6
t = 3
√(x-1) = 3
x-1 = 9
x = 10
Ответ: x∈[5;10] - бесконечное число решений (Д)