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Ответ

Ответ дан
Nullpo201

2)|x|(|x|-5) = 0

|x| = 0 => x = 0

|x| = 5 => x = ±5

3)|x²-x-1| = 1

[x²-x-1 = 1

[x²-x-1 = -1

1. x²-x-2 = 0

Теорема Виета:

{x1+x2 = 1

{x1•x2 = -2

x1 = -1

x2 = 2

2. x²-x = 0

x(x-1) = 0

x = 0

x = 1

4)|x²+5| = 6x

x²+5 при любом x положительный, знак модуля можно убрать

x²+5 = 6x

x²-6x+5 = 0

Теорема Виета:

{x1+x2 = 6

{x1•x2 = 5

x1 = 1

x2 = 5

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