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valveul99
@valveul99
April 2022
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Помогите пожалуйста найти такие x, чтобы числа 3x+2, x-1, 4x+3 были членами арифметической прогрессии.
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ATLAS
Verified answer
D=(x-1)-(3x+2)=x-1-3x-2=-2x-3
d=(4x+3)-(x-1)=4x+3-x+1=3x+4 => 3x+4=-2x-3
3x+2x=-4-3
5x=-7
x=-7/5
x=-1,4
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valveul99
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Answers & Comments
Verified answer
D=(x-1)-(3x+2)=x-1-3x-2=-2x-3d=(4x+3)-(x-1)=4x+3-x+1=3x+4 => 3x+4=-2x-3
3x+2x=-4-3
5x=-7
x=-7/5
x=-1,4