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@kkkkiiiii
August 2022
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При каких значениях параметра a уравнение x3-3x2+6=a имеет два корня?
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sangers1959
Verified answer
X³-3x²+6=a
a=6
x³-3x²+6=6
x³-3x²=0
x²*(x-3)=0
x²=0
x₁=0
x-3=0
x₂=3.
Ответ: при a=6 x₁=0, x₂=3.
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Answers & Comments
Verified answer
X³-3x²+6=aa=6
x³-3x²+6=6
x³-3x²=0
x²*(x-3)=0
x²=0
x₁=0
x-3=0
x₂=3.
Ответ: при a=6 x₁=0, x₂=3.