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iluxandro
@iluxandro
July 2022
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Решите биквадратное уравнение, пожалуйста.
4x^4-3x^3-8x^2+3x+4=0
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hraky
(2x^2-2)^2-3x(x^2-1)=0
(x^2-1)(4(x^2-1)-3x)=0
x1=-1
x2=1
4x^2-3x-4=0
D=9+64=73
x3=(3-корень(73))/8
x4=(3+корень(73))/8
Теперь вроде правильно)
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Answers & Comments
(x^2-1)(4(x^2-1)-3x)=0
x1=-1
x2=1
4x^2-3x-4=0
D=9+64=73
x3=(3-корень(73))/8
x4=(3+корень(73))/8
Теперь вроде правильно)