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nicesmall
@nicesmall
July 2022
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Помогите, модули, срочно
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sangers1959
Verified answer
|x²-4x+3|=2x-2
д)
|x²-x-3x+3|=2(x-1)
|x(x-1)-3(x-1)||=2(x-1)
|(x-1)(x-3)|-2(x-1)=0
Раскрываем модуль и получаем систему уравнений:
(x-1)(x-3)-2(x-1)=0 (x-1)(x-3-2)=0 (x-1)(x-5)=0 x-1=0 x₁=1 x-5=0 x₂=5
(x-1)(3-x)-2(x-1)=0 (x-1)(3-x-2)=0 (x-1)(1-x)=0 x-1=0 x₃=1 1-x=0 x₄=1
Ответ: x₁=1 x₂=5.
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Answers & Comments
Verified answer
|x²-4x+3|=2x-2д)
|x²-x-3x+3|=2(x-1)
|x(x-1)-3(x-1)||=2(x-1)
|(x-1)(x-3)|-2(x-1)=0
Раскрываем модуль и получаем систему уравнений:
(x-1)(x-3)-2(x-1)=0 (x-1)(x-3-2)=0 (x-1)(x-5)=0 x-1=0 x₁=1 x-5=0 x₂=5
(x-1)(3-x)-2(x-1)=0 (x-1)(3-x-2)=0 (x-1)(1-x)=0 x-1=0 x₃=1 1-x=0 x₄=1
Ответ: x₁=1 x₂=5.