Verified answer

одз вспомним

1. 2 + x >= 0

x >= -2

√(x + 2)² = (√(2 + x))²

|x + 2| = x + 2

при x >= -2    |x + 2| = x + 2

x + 2 = x + 2  

x ∈ [-2, +∞)

2. √((x - 2)(x - 5)) = √(2 - x) * √(5 - x)

одз (x - 2)(x - 5) ≥ 0

++++++++[2] ------------ [5] ++++++++

x ∈ (-∞, 2] U [5, +∞)

2 - x >=0

x <= 2

5 - x >=0

x <= 5

x ∈ (-∞, 2]

√(x - 2)(x - 5) = √(2 - x)* √(5 - x)

√(x - 2)(x - 5)² = √(2 - x)²* √(5 - x)²

(x - 2)(x - 5) - (2 - x)* (5 - x) = 0

x <= 2

x = 5 нет

ответ х <= 2