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Zahid07
@Zahid07
August 2022
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x^3-5x+4=0 ? помогите.
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moboqe
X=1(делитель свободного члена
4
)
1^3-5*1+4=0
0=0⇒ x=1 -- корень уравнения
x^3-5x+4 |_x-1____
x^3-x^2 x^2+x-4
-----------
x^2-5x
x^2-x
----------
-4x+4
-4x+4
-------
0
x^3-5x+4=0
(x-1)(
x^2+x-4)=0
x1=1
x^2+x-4=0
D=1+16=17
x2=(-1+sqrt(17))/2
x3=(-1-sqrt(17))/2
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Zahid07
А можно методом разложения на множители?
Zahid07
По теореме Безу не то.
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Answers & Comments
1^3-5*1+4=0
0=0⇒ x=1 -- корень уравнения
x^3-5x+4 |_x-1____
x^3-x^2 x^2+x-4
-----------
x^2-5x
x^2-x
----------
-4x+4
-4x+4
-------
0
x^3-5x+4=0
(x-1)(x^2+x-4)=0
x1=1
x^2+x-4=0
D=1+16=17
x2=(-1+sqrt(17))/2
x3=(-1-sqrt(17))/2