Предмет: Алгебра, автор: qwerty7632

Решите неравенство.
\sqrt{x + 6} < x
\sqrt{3 + 2x} \geqslant \sqrt{x + 1}
\sqrt{ {x}^{2} - 4x} > x - 3
​

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Автор ответа: QazMlp17
1
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