Решите неравенство: Cos x ≥ x² + 1 Желательно подробно.. Created by lyrew. algebra-ru

task/25016556-------------------Решите неравенство: cos(x) ≥x²+1-----------------решение :Левая часть неравенства не больше 1 (функция f(x)= cos(x) ограниченная: -1≤ cos(x) ≤ 1, max (cosx) = 1 ,если x=2πn,n∈Z) ,aправая часть не меньше 1, min (x² + 1) =1 ,если x=0 .Значит cos(x) ≥x²+1⇔{cos(x)=1 ; x²+1=1.⇔{x=2πn,n∈Z ; x²+1=1. ⇒x=0.ответ : x=0.Новые вопросы в АлгебраПомогите с интеграломПомогите с интегралом пожалуйстаЗнайдіть розв’язок нерівності: 3х − 6 < 3.Довести нерівність 3x/(9x²+1)≤1/28. Знайдіть всі значення х, при яких має зміст вираз:1/(x + 1)1. Розв'язати нерівність розв'язками нерівності. 1/(2(x + 1) - 1/2x <= 1 У вiдповiдь записати суму всіх цілих чисел, які не2. Розв'язати нерівніс…ть розв'язками нерівності. 1/x² - 3x + 2 <= 1/2 У відповідь записати добуток усіх цілих чисел, які не3. Розв'язати нерівність розв'язок. x - 8/x² - 5x + 4 > 2/x + 1. У відповідь записати найменший додатний цілиAB=√(Xb-Xa)2+(Yb-Ya)2 как читается формулаУ шкільній їдальні учнів можна розсадити по чотири за столом. Яку найменшу кількість столівзаймуть учні першого класу, якщо у ньому навчається 25 учн…ів?Пожалуйста помогите!Х + 5/12= 11/12Плиз Пожалуйста помогите!Х + 5/12= 11/12Срочно ооооПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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\u0420\u043e\u0437\u0432'\u044f\u0437\u0430\u0442\u0438 \u043d\u0435\u0440\u0456\u0432\u043d\u0456\u0441\u0442\u044c \u0440\u043e\u0437\u0432'\u044f\u0437\u043a\u0430\u043c\u0438 \u043d\u0435\u0440\u0456\u0432\u043d\u043e\u0441\u0442\u0456. 1\/(\n2(x + 1) - 1\/2x <= 1 \u0423 \u0432i\u0434\u043f\u043e\u0432i\u0434\u044c \u0437\u0430\u043f\u0438\u0441\u0430\u0442\u0438 \u0441\u0443\u043c\u0443 \u0432\u0441\u0456\u0445 \u0446\u0456\u043b\u0438\u0445 \u0447\u0438\u0441\u0435\u043b, \u044f\u043a\u0456 \u043d\u0435\n2. \u0420\u043e\u0437\u0432'\u044f\u0437\u0430\u0442\u0438 \u043d\u0435\u0440\u0456\u0432\u043d\u0456\u0441\u0442\u044c \u0440\u043e\u0437\u0432'\u044f\u0437\u043a\u0430\u043c\u0438 \u043d\u0435\u0440\u0456\u0432\u043d\u043e\u0441\u0442\u0456. 1\/x\u00b2 - 3x + 2 <= 1\/2 \u0423 \u0432\u0456\u0434\u043f\u043e\u0432\u0456\u0434\u044c \u0437\u0430\u043f\u0438\u0441\u0430\u0442\u0438 \u0434\u043e\u0431\u0443\u0442\u043e\u043a \u0443\u0441\u0456\u0445 \u0446\u0456\u043b\u0438\u0445 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Решите неравенство: Cos x ≥ x² + 1 Желательно подробно...

lyrew @lyrew

July 2022 1 5 Report

Решите неравенство:
Cos x ≥ x² + 1
Желательно подробно.

Answers & Comments

oganesbagoyan

Verified answer

task/25016556
-------------------
Решите неравенство: cos(x) ≥x²+1
-----------------
решение :
Левая часть неравенства не больше 1 (функция f(x)= cos(x) ограниченная: -1≤ cos(x) ≤ 1, max (cosx) = 1 ,если x=2πn,n∈Z) ,a
правая часть не меньше 1, min (x² + 1) =1 ,если x=0 .
Значит cos(x) ≥x²+1⇔{cos(x)=1 ; x²+1=1.⇔{x=2πn,n∈Z ; x²+1=1.
⇒x=0.

ответ : x=0.