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m369mm
@m369mm
July 2022
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Найдите сумму корней уравнения
Х^2-2х=/х-1/ пожалуй помогите
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oOlDarknesslOo
X²-2x=|x-1| ⇒ {1. x-1=x²-2x ∧ x≥0} ∨ {2. x-1=-(x²-2x) ∧ x<0}
1. x-1=x²-2x ⇒ x²-3x+1=0.
x²-3x+1=0;
a=1; b=-3; c=1;
D=b²-4ac=9-4=5
x1,2=(-b±√D)/2a=(3±√5)/2;
{x1>0 ∨ x2>0} ∧ x≥0 ⇒ x1+x2=-b/a=3.
2. x-1=-(x²-2x) ⇒ x²-x-1=0.
x²-x-1=0;
a=1; b=-1; c=1;
D=b²-4ac=1+4=5;
x3,4=(-b±√D)/2a=(1±√5)/2;
{x3>0 ∨ x4<0} ∧ x<0 ⇒ x4=(1-√5)/2.
x1+x2+x4=3+(1-√5)/2=(7-√5)/2.
Ответ: (7-√5)/2.
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Answers & Comments
1. x-1=x²-2x ⇒ x²-3x+1=0.
x²-3x+1=0;
a=1; b=-3; c=1;
D=b²-4ac=9-4=5
x1,2=(-b±√D)/2a=(3±√5)/2;
{x1>0 ∨ x2>0} ∧ x≥0 ⇒ x1+x2=-b/a=3.
2. x-1=-(x²-2x) ⇒ x²-x-1=0.
x²-x-1=0;
a=1; b=-1; c=1;
D=b²-4ac=1+4=5;
x3,4=(-b±√D)/2a=(1±√5)/2;
{x3>0 ∨ x4<0} ∧ x<0 ⇒ x4=(1-√5)/2.
x1+x2+x4=3+(1-√5)/2=(7-√5)/2.
Ответ: (7-√5)/2.