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seereejkee
@seereejkee
August 2022
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люди, производную пож,
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manyny06
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Решение смотри на фотографии
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President01
(x^2+3x/x-1)' =
= [(x^2+3x)' * (x-1) - (x^2+3x) * (x-1)'] / ([-1)^2 =
= [2x+3(x-1) - x^2-3x] / (x-1)^2 =
= (2x+3x-3-x^2-3x) / (x^2-2x+1) =
= - (x^2-2x+3) / x^2-2x+1
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seereejkee
пасаны еще диф. уравнения: у=у"+у'-6у=0; у=3; у'=1; при х=0
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Verified answer
Решение смотри на фотографии= [(x^2+3x)' * (x-1) - (x^2+3x) * (x-1)'] / ([-1)^2 =
= [2x+3(x-1) - x^2-3x] / (x-1)^2 =
= (2x+3x-3-x^2-3x) / (x^2-2x+1) =
= - (x^2-2x+3) / x^2-2x+1