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maksimfilipov
@maksimfilipov
July 2022
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найдите произведение корней уравнения (х^2-2х)^2-3х^2+6х-4=0
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m11m
(x² -2x)² -3x²+6x-4=0
(x² -2x)² -(3x² -6x) -4=0
(x² -2x)² -3(x² -2x) -4=0
Пусть y=x² -2x
y² -3y-4=0
D=9+16=25
y₁=
3-5
= -1
2
y₂ =
3+5
=4
2
При у=-1
x² -2x=-1
x² -2x+1=0
(x-1)² =0
x=1
При у=4
x² -2x-4=0
D=4+16=20
x₁=
2-√20
=
2-2√5
= 1-√5
2 2
x₂ =1+√5
Произведение корней:
1 * (1-√5) * (1+√5) = 1 - (√5)² = 1-5=-4
Ответ: -4.
1 votes
Thanks 1
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Answers & Comments
(x² -2x)² -(3x² -6x) -4=0
(x² -2x)² -3(x² -2x) -4=0
Пусть y=x² -2x
y² -3y-4=0
D=9+16=25
y₁=3-5= -1
2
y₂ =3+5 =4
2
При у=-1
x² -2x=-1
x² -2x+1=0
(x-1)² =0
x=1
При у=4
x² -2x-4=0
D=4+16=20
x₁=2-√20 = 2-2√5 = 1-√5
2 2
x₂ =1+√5
Произведение корней:
1 * (1-√5) * (1+√5) = 1 - (√5)² = 1-5=-4
Ответ: -4.