October 2021 1 31 Report
50 баллов! Срочно! Решить неравенства! С подробным решением!
[tex] \sqrt{ {x}^{2} + 1 } > - 2[/tex]
[tex] \sqrt{x + 1} < - 2[/tex]
[tex] \frac{1}{ \sqrt{3 - x} } > 0[/tex]
[tex] \sqrt{x} > \sqrt{2x - 3} [/tex]
Please enter comments
Please enter your name.
Please enter the correct email address.
You must agree before submitting.

Answers & Comments


Copyright © 2024 SCHOLAR.TIPS - All rights reserved.