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smilee12345
@smilee12345
August 2022
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(x^3–3)^2+7(x^3–3)–4=0 Помогите решить пожалуйст
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АндрейРедченко
(x^3 - 3)^2 + 7(x^3 - 3) - 4 = 0
a = 1; b = 7; c = -4
D = b^2 - 4ac
D = 7^2 - 4*(-4) = 49 + 16 = 65
x1 = (-7 + k65)/2
x1 = x^3 - 3
(-7 + k65)/2 = x^3 -3
-7 + k65 = 2x^3 -3
2x^3=k65 - 4
x = K((k65-4)/2)
x2 = (-7 - k65)/2
x2 = x^3 - 3
(-7 - k65)/2 = x^3 -3
-7 - k65 = 2x^3 -3
2x^3= -k65 - 4
x = K((-k65-4)/2)
Ответ: K((+- k65 -4)/2)
k - квадратный корень
K - кубический корень
+- - плюс-минус
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АндрейРедченко
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Answers & Comments
a = 1; b = 7; c = -4
D = b^2 - 4ac
D = 7^2 - 4*(-4) = 49 + 16 = 65
x1 = (-7 + k65)/2
x1 = x^3 - 3
(-7 + k65)/2 = x^3 -3
-7 + k65 = 2x^3 -3
2x^3=k65 - 4
x = K((k65-4)/2)
x2 = (-7 - k65)/2
x2 = x^3 - 3
(-7 - k65)/2 = x^3 -3
-7 - k65 = 2x^3 -3
2x^3= -k65 - 4
x = K((-k65-4)/2)
Ответ: K((+- k65 -4)/2)
k - квадратный корень
K - кубический корень
+- - плюс-минус