Найти значения а, при которых решение уравнения сужествуют и принадлежат отрезку 2; 17

√(x + 3 - 4√(x - 1)) + √(x + 8 - 6√(x - 1)) = a

√(x - 1 - 2·√(x - 1)·2 + 4) + √(x - 1 - 2·√(x - 1)·3 + 9) = a

√(√(x - 1) - 2)² + √(√(x - 1) - 3)² = a

|√(x - 1) - 2| + |√(x - 1) - 3| = a

0 ≤ √(x - 1) < 2, 2 - √(x - 1) + 3 - √(x - 1) = a, x ∈ [2; 17]

2√(x - 1) = 5 - a, 0 ≤ x - 1 < 4, x ∈ [2; 17]

√(x - 1) = (5 - a)/2, 1 ≤ x < 5, x ∈ [2; 17]

x - 1 = (5 - a)²/4, 2 ≤ x < 5, a ≤ 5

x = (5 - a)²/4 + 1, 2 ≤ x < 5, a ≤ 5

2 ≤ (5 - a)²/4 + 1 < 5, a ≤ 5

1 ≤ (5 - a)²/4 < 4, a ≤ 5

1 ≤ (5 - a)/2 < 2, a ≤ 5

2 ≤ 5 - a < 4, a ≤ 5

-3 ≤ - a < -1, a ≤ 5

1 < a ≤ 3, a ≤ 5

1 < a ≤ 3

2 ≤ √(x - 1) < 3, √(x - 1) - 2 + 3 - √(x - 1) = a, x ∈ [2; 17]

4 ≤ x - 1 < 9, a = 1, x ∈ [2; 17]

5 ≤ x < 10, a = 1, x ∈ [2; 17]

5 ≤ x < 10, a = 1

a = 1

√(x - 1) ≥ 3, √(x - 1) - 2 + √(x - 1) - 3 = a, x ∈ [2; 17]

√(x - 1) ≥ 3, 2√(x - 1) = a + 5, x ∈ [2; 17]

√(x - 1) ≥ 3, √(x - 1) = (a + 5)/2, x ∈ [2; 17]

x - 1 ≥ 9, x - 1 = (a + 5)²/4, x ∈ [2; 17], a + 5 > 0

x ≥ 10, x = (a + 5)²/4 + 1, x ∈ [2; 17], a + 5 > 0

x = (a + 5)²/4 + 1, x ∈ [10; 17], a + 5 > 0

10 ≤ (a + 5)²/4 + 1 ≤ 17, a + 5 > 0

9 ≤ (a + 5)²/4 ≤ 16, a + 5 > 0

3 ≤ (a + 5)/2 ≤ 4, a + 5 > 0

6 ≤ a + 5 ≤ 8, a + 5 > 0

6 ≤ a + 5 ≤ 8

1 ≤ a ≤ 3

ответ: a ∈ [1; 3].