Вопрос задан 10.07.2023 в 06:56. Предмет Алгебра. Спрашивает Гриценко Арсентий.

Дана система уравнений: {2a+b=15 −b+a=0 Вычисли значение переменной a.

Смотреть ответ kocherijkapiva kocherijkapiva Ответ: {2a+b=15 −b+a=0 {2a+b=15 a-b=0 3a=15 a=5 Новые вопросы в Алгебра СРОЧНО!!! Приведите выражение к многочлену стандартного вида, используя формулы сокращённого умножения:(1/2m+4n)^2 5. Подайте тричлен у чигляді У зигляді квадрата двочлена: квадрата двочлена: а² + 8 +16.​ ДУЖЕ ТЕРМІНОВО! Розв'язати систему рiвнянь ​ СРОЧНО!!! Докажите, что при любом целом значении х значение выражения делится на 10. (3x-1)(3x+1)+(x-4)^2-2x+5 ПОМОГИТЕ ПОЖАЛУЙСТА С АЛГЕБРОЙ, СРОЧНО Предыдущий Следующий Вместе мы знаем больше Этот сайт использует cookies Политика Cookies . Вы можете указать условия хранения и доступ к cookies в своем браузере. 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Вы задали несколько вопросов. Давайте по порядку рассмотрим каждый из них:

1. Приведение выражения к многочлену стандартного вида: Имеется выражение (1/2m + 4n)^2. Для приведения его к стандартному виду, воспользуемся формулой квадрата бинома:

(a + b)^2 = a^2 + 2ab + b^2

В данном случае, a = 1/2m, b = 4n. Подставим значения:

(1/2m + 4n)^2 = (1/2m)^2 + 2 * (1/2m) * (4n) + (4n)^2 = 1/4m^2 + 4mn + 16n^2

2. Тричлен в виде квадрата двучлена: У вас дано выражение a² + 8 + 16. Заметим, что 8 и 16 являются квадратами чисел, и похоже, что вы хотите представить это выражение в виде квадрата двучлена. Воспользуемся формулой квадрата двучлена:

(a + b)^2 = a^2 + 2ab + b^2

Если a^2 = a², то 8 = 2ab и 16 = b². Решим систему уравнений:

  1. a^2 = a² → a = 0 (потому что квадрат числа не может быть отрицательным)
  2. 8 = 2ab → 8 = 2 * 0 * b → b = любое число

Таким образом, данное выражение не может быть представлено в виде квадрата двучлена.

3. Решение системы уравнений: У вас не указана система уравнений. Пожалуйста, предоставьте систему уравнений, которую вы хотите решить.

4. Доказательство деления выражения на 10: У вас дано выражение (3x - 1)(3x + 1) + (x - 4)^2 - 2x + 5. Чтобы доказать, что оно делится на 10 при любом целом значении x, следует показать, что оно имеет остаток 0 при делении на 10.

Подставим выражение и упростим его: (3x - 1)(3x + 1) + (x - 4)^2 - 2x + 5 = 9x^2 - 1 + x^2 - 8x + 16 - 2x + 5 = 10x^2 - 10x + 20

Как видно, второе слагаемое 10x^2 - 10x + 20 делится на 10 без остатка для любого целого значения x. Таким образом, данное выражение действительно делится на 10 при любом целом значении x.

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