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DenStiven
@DenStiven
July 2022
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Помогите решить.
4+12х≥7+13х
И второе
-(4-х)≤2(3+х)
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АлександринA
12x-13x≥-4+7
-x≥3(делим
на минус 1)
x
≤-3
x-2x
≤4+6
-x
≤10(делим на минус 1)
x≥-10
1 votes
Thanks 2
mpimenov2015
1. 4+12x
7+13x
4-7
13x-12x
-3
x
x
-3
2. -(4-x)
2(3+x)
-4+x
6+2x
-4-6
2x-x
-10
x
x
-10
1 votes
Thanks 2
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Answers & Comments
-x≥3(делим на минус 1)
x≤-3
x-2x≤4+6
-x≤10(делим на минус 1)
x≥-10
4-713x-12x
-3x
x-3
2. -(4-x)2(3+x)
-4+x6+2x
-4-62x-x
-10x
x-10