Question

$\sqrt{ {x}^{2} - 7x + 5} \geqslant \sqrt{3x - 4}$
решите неровность​

Answer 1

√f(x) ≥ √g(x) ⇔ f(x) ≥ g(x)   g(x) ≥ 0

√(x² - 7x + 5) ≥ √(3x - 4)

3x - 4 ≥ 0

x² - 7x + 5 ≥ 3x - 4

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x ≥ 4/3

x² - 10x + 9 ≥ 0

(x - 9)(x - 1) ≥ 0

+++++++++[1] -------------- [9] ++++++++

x ∈ (-∞, 1] U [9, +∞) + x ≥ 4/3

x ∈ [9, +∞)