4х – 5 3x + 1 3x + 7 x + 15 4 + 12х 7 + 13х 7 – 4х - вопрос №4531371541271374616802365 от бека151 03.02.2021 06:20

Question

Математика: 4х – 5 3x + 1 3x + 7 x + 15 4 + 12х 7 + 13х 7 – 4х 6х – 23 4∙(x – 3) + 5x 3x 2∙(3x + 1) – x 3∙(x + 4) 7x + 4∙(x – 2) 6∙(1 + 3x) 8 + 5х ≤ 3∙(7 + 2х)

бека151 · Answer

ID1845503354 the useful lang the usefu is a 1ti_ to be the useful

Пошаговое объяснение:

ID1845503354 the useful lang the useful lang the useful lang and I LOVE PLAYING FREE FIRE in a 1ti_ the usefu is not a mong about it and it will have the

ID1845503354 the useful lang the useful lang the useful for the usefu out the useful lang the useful lang the useful lang and

4х – 5 < 3x + 1 3x + 7 > x + 15 4 + 12х > 7 + 13х 7 – 4х < 6х – 23 4∙(x – 3) + 5x 3x 2∙(3x + 1) – x 3∙(x + 4) 7x + 4∙(x – 2) > 6∙(1 + 3x) 8 + 5х ≤ 3∙(7 + 2х)

4х – 5 < 3x + 1 3x + 7 > x + 15
4 + 12х > 7 + 13х
7 – 4х < 6х – 23
4∙(x – 3) + 5x 3x
2∙(3x + 1) – x 3∙(x + 4)
7x + 4∙(x – 2) > 6∙(1 + 3x)
8 + 5х ≤ 3∙(7 + 2х)