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Hikit123
@Hikit123
July 2022
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Решите неравенство методом интервалов а) 3|х+2|+|х-2|<4(х+3)
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sedinalana
Verified answer
1)x<-2
-3x-6-x+2<4x+12
-4x-4x<12+4
-8x<16
x>-2
нет решения
2)-2≤x≤2
3x+6-x+2<4x+12
2x-4x<12-8
-2x<4
x>-2
-2<x≤2
3)x>2
3x+6+x-2<4x+12
0<8
нет решения
Ответ x∈(-2;2]
0 votes
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Answers & Comments
Verified answer
1)x<-2-3x-6-x+2<4x+12
-4x-4x<12+4
-8x<16
x>-2
нет решения
2)-2≤x≤2
3x+6-x+2<4x+12
2x-4x<12-8
-2x<4
x>-2
-2<x≤2
3)x>2
3x+6+x-2<4x+12
0<8
нет решения
Ответ x∈(-2;2]