Question

Решить неравенство методом интервалов

Срочно ✔️✔️✔️✔️✔️✔️✔️✔️

Answer 1

Ответ:

$20)\ \ 2x^2-|x|-1\geq 0\ \ \ \\\\x^2=|x|^2\ \ \ \Rightarrow \ \ \ 2|x|^2-|x|-1\geq 0\\\\t=|x|\geq 0\ \ ,\ \ \ 2t^2-t-1\geq 0\ \ ,\ \ D=9\ \ ,\ \ t_1=-\dfrac{1}{2}\ \ ,\ \ t_2=1\\\\2(t+\frac{1}{2})(t-1)\geq 0\ \ ,\ \ \ znaki:\ \ +++[-\frac{1}{2}\ ]---[\ 0\ ]---[\ 1\ ]+++\\\\t\in [\ 1\ ;\ +\infty )\ \ \ \Rightarrow \ \ \ |x|\geq 1\ \ \Rightarrow \ \ \ \left[\begin{array}{l}x\geq 1\\x\leq -1\end{array}\right\\\\\\x\in (-\infty \, ;\,\ -1\ ]\cup [\ 1\ ;+\infty \, )$