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√(x²+4x+4) ≥ 0
x²+4x+4 = (x+2)²
(x+2)² ≥ 0 zawsze !!!
√(x²+4x+4) <=> |x+2|
|x+2| ≥ 0
x ∈ (-∞;+∞) lub x ∈ R
2.
11 - √(x²-6x+9)
x²-6x+9 = (x-3)²
(x-3)² ≥ 0 zawsze !!!
x ∈ (-∞;+∞) lub x ∈ R
11 - √(x²-6x+9) <=> 11 -|x-3|
11 -|x-3|≥ 0
dla x ∈ (-∞;3)
11 + x-3 ≥ 0
x ≥ -8
dla x ∈ (3;+∞)
11 - x+3 ≥ 0
x ≤ 14
x ∈ [-8;14]
11 -|x-3|< 0
x ∈ R/[-8;14] => x ∈ (-∞;-8)(14;+∞)