Answer 1

Решение:
$\sqrt{5x - 2} \ \textless \ 4$
$\left \{ {{5x- 2 \ \textless \ 16} \atop {5x- 2 \geq 0}} \right.$
$0 \leq 5x- 2 \ \textless \ 16 \\ 0+ 2 \leq 5x \ \textless \ 16 + 2 \\ 2 \leq 5x \ \textless \ 18$
$2: 5 \leq x \ \textless \ 18 : 5 \\ 0,4 \leq x \ \textless \ 3,6$
х ∈ [0,4; 3,6)
Ответ: [0,4; 3,6)

Answer 2

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