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miham7048
@miham7048
July 2022
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решите с пояснениями пж
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kisaaa73
(2х+2)/(х-7)≥1 х≠7
(2х+2)/(х-7) - 1 ≥0
(2х+2-х+7)/х-7≥0
(х+2+7)/(х-7)≥0
(х+9)/(х-7)≥0
х≥-9
х<7
х€(-7;+∞)U(-∞;-9]
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Answers & Comments
(2х+2)/(х-7) - 1 ≥0
(2х+2-х+7)/х-7≥0
(х+2+7)/(х-7)≥0
(х+9)/(х-7)≥0
х≥-9
х<7
х€(-7;+∞)U(-∞;-9]