Question

Найдите все значения а при каждом из которых уравнение имеет один корень х^2+(1-a)^2=|x-1+a|+|x-a+1|

Answer 1

Введем новую переменную t=a-1,
x^2+t^2=|x+t|+|x-t|
построим графики функций у=x^2+t^2; y=|x-t|; y=|x+t|
по оси  х отмечаем  -t I  t; по оси у;  t^2;|t|
строим параболу( вершина (0:t^2)   и  |x|; -|x| -то углы с вершиной (-t;0) i (t;0)
t^2=2|t|
t>0; t^2=2t; t^2-2t=0; t(t-2)=0; t=0  ili  t=2
 a-1=2; a=3
  если t=0,
x^2+(1-1)^2=|(x-1+1|+|x-1+1|
x^2=2|x| два корня, а=1-не подходит

 t<0;   t^2=-2t; t^2+2t=0; t=0  ili t=-2
a-1=-2; a=-1
Ответ. -1;3

,

Answer 2

Данное уравнение имеет один корень при следующих значениях параметра a:

1. a = 1: Подставим a = 1 в уравнение: x^2 + 0 = |x| + |x| x^2 = 2|x| x^2 = 2x, x >= 0 x^2 - 2x = 0 x(x - 2) = 0 Таким образом, уравнение имеет один корень x = 0.

2. a = 0: Подставим a = 0 в уравнение: x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| x^2 + 1 = |x-1| + |x+1| 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