Question

помогите решить пожалуйста
срочно



$\sqrt{3x ^{2} - 18x + 25} + \sqrt{4x ^{2} - 24x + 29 } = 6x - x ^{2} - 4$
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Answer 1

√(3x² - 18x + 25) + √(4x² - 24x + 29) = 6x - x² - 4

3x² - 18x + 25 ≥ 0

D = 18² - 4*3*25 = 24

x12 = (18 +- √24)/6 = 3 +- √6/6

x ∈ (-∞, 3 - √6/6] U [ 3 + √6/6, +∞)

4x² - 24x + 29 ≥ 0

D = 24² - 4*4*29 = 112

x12 = (24 +- √112)/8 = 3 +- √7/2

x ∈ (-∞, 3 - √7/2] U [ 3 + √7/2, +∞)

6x - x² - 4 ≥ 0

x² - 6x + 4 ≤ 0

D = 36 - 16 = 20

x12 = (6 +- √20)/2 = 3 +- √5

x ∈ [3 - √5, 3 + √5]

x ∈ [3 - √5, 3 - √7/2] U [3 - √7/2, 3 + √5]

замена

√(3x² - 18x + 25) = a

√(4x² - 24x + 29) = b

a + b = a² - b²

a + b = (a - b)(a + b)

(a + b)(a - b - 1) = 0

a + b = 0

√(3x² - 18x + 25) = -√(4x² - 24x + 29)

корни не совпадают x ∈ ∅

a - b = 1

√(3x² - 18x + 25) = 1 + √(4x² - 24x + 29)

(3x² - 18x + 25) = 1 + 2√(4x² - 24x + 29) + (4x² - 24x + 29)

6x - x² - 4 = 1 + 2√(4(x² - 6x + 4)+13)

x² - 6x + 4 = y

-y = 1 + 2√(4y + 13)

-y - 1 = 2√(4y + 13)

y < = -1

y² + 2y + 1 = 4(4y + 13)

y² + 2y + 1 = 16y + 52

y² - 14y - 51 = 0

D = 14² + 4*51 = 400

y12 = (14 +- 20)/2 = -3    17

17 > -1 нет

y = -3

x² - 6x + 4 = -3

x² - 6x + 7 = 0

D = 36 - 28 = 0

x12 = (6 +- √8)/2 = (6 +- 2√2)/2 = 3 +- √2

ответ  3 +- √2

фу ..... наконец