Помогите пожалуйста, заранее спасибо. Created by way51. algebra-ru

ОДЗ: 7-x²≥0 и -6x≥0 x² -7≤0 x≤0 (x-√7)(x+√7)≤0 + - +---------- -√7 ------------ √7 ------------- \\\\\\\\\\\\\\\\\\ x∈[-√7; √7]{x∈[-√7; √7]{x≤0 ⇒ x∈[-√7; 0]7-x²= -6x-x²+6x+7=0x²-6x-7=0D=(-6)² -4*(-7)=36+28=64=8²x₁=(6-8)/2= -2/2= -1x₂=(6+8)/2=14/2=7∉[-√7; 0] - не корень уравнения.Проверка корня:х= -1√(7-1)=√(-6*(-1))√6=√6 - верноОтвет: -1.Pacanka202Pacanka202Возведём в квадрат обе части уравнения:7–х^2=–6хх^2–6х–7=0Д=/36–4•1•(-7)=/64=8х1=(6+8)/2=7х2=(6–8)/2=–1Новые вопросы в АлгебраСрочно нужно решить через дискриминант 16х^2 - 8х + 1 = 0 пж пж пж Помогите срочно нужно решить пожалуйста, задание на фотоПомогите срочно, задание на фото,тут тоже уже другое заданиеПомогите срочно, задание на фото,тут уже другоеПомогите срочно, задание на фото214. Моторний човен проплив 8 км за течією річки й повернувся назад, витративши на весь шлях 54 хв. Знайдіть швидкість течії річки, якщо власна швидкі…сть човна становить 18 км/год.Решите пожалуйста с подробным объяснением.Поясніть будь ласка простими словами (бажано з прикладами) що таке теорія ймовірності і як її використовувати!(-0.2)^5 : (0.1)^4Сколько будет помогитеСкільки площин симетрії має правильна трикутна призма?ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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Помогите пожалуйста, заранее спасибо...

way51 @way51

July 2022 1 9 Report

Помогите пожалуйста, заранее спасибо

Answers & Comments

Pacanka202 ОДЗ: 7-x²≥0 и -6x≥0
x² -7≤0 x≤0
(x-√7)(x+√7)≤0
+ - +
---------- -√7 ------------ √7 -------------
\\\\\\\\\\\\\\\\\\
x∈[-√7; √7]

{x∈[-√7; √7]
{x≤0 ⇒ x∈[-√7; 0]

7-x²= -6x
-x²+6x+7=0
x²-6x-7=0
D=(-6)² -4*(-7)=36+28=64=8²
x₁=(6-8)/2= -2/2= -1
x₂=(6+8)/2=14/2=7∉[-√7; 0] - не корень уравнения.

Проверка корня:
х= -1
√(7-1)=√(-6*(-1))
√6=√6 - верно

Ответ: -1.