Предмет: Алгебра, автор: Аноним

Алгебра. Даю 10 балів.

Виконати 1 завдання.

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Автор ответа: IUV
1

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IUV: на сьогодні досить
Аноним: добре) дякую)
Автор ответа: NNNLLL54
0

Ответ:

Правило :

если   \bf |\, x\, |\leq a  ,   то   \bf -a\leq x\leq a  .

\bf |x^2-2|\leq x\ \ \ \Rightarrow \ \ \ -x\leq x^2-2\leq x\ \ ,\\\\\left\{\begin{array}{l}\bf x^2-2\leq x\\\bf x^2-2\geq -x\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2-x-2\leq 0\\\bf x^2+x-2\geq 0\end{array}\right\ \ \left\{\begin{array}{l}\bf (x-2)(x+1)\leq 0\\\bf (x+2)(x-1)\geq 0\end{array}\right\\\\\\\left\{\begin{array}{l}\bf x\in [-1\ ;\ 2\ ]\\\bf x\in (-\infty ;-2\ ]\cup [\ 1\ ;+\infty \, )\end{array}\right\ \ \Rightarrow \ \ \ x\ \in [\ 1\ ;\ 2\ ]  

Ответ:    \bf x\in [\ 1\ ;\ 2\ ]\ .  

P.S.   Решение квадратных неравенств .

По теореме Виета находим корни квадратных трёхчленов и решаем неравенства методом интервалов :

\bf a)\ \ x^2-x-2=0\ ,\ \ x_1=-1\ ,\ x_2=2\ \ \ \Rightarrow \ \ \ x^2-x-2=(x-2)(x+1)\ ,\\\\(x-2)(x+1)\leq 0\ \ ,\ \ \ znaki         :\ +++[-1\, ]---[\ 2\ ]+++\\\\x\in [-1\ ;\ 2\ ]\\\\b)\ \  x^2+x-2=0\ ,\ \ x_1=-2\ ,\ x_2=1\ \ \ \Rightarrow \ \ \ x^2+x-2=(x+2)(x-1)\ ,\\\\(x+2)(x-1)\geq 0\ \ ,\ \ \ znaki         :\ +++[-2\, ]---[\ 1\ ]+++\\\\x\in (-\infty\ ;\, -2\ ]\cup [\ 1\ ;+\infty \, )  

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