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Решите систему уравнений: A) {2u+5v=0, {-8u+15v=7; Смотреть ответ ildar502020 ildar502020 Ответ: Объяснение: Решите систему уравнений: {2u+5v=0;    [*4]           (1) {-8u+15v=7;                 (2) ----------------- 8u+20v=0;                  (3) Складываем (2)  и (3): 35v=7; v=1/5    подставляем в (1) 2u+5*(1/5)=0; 2u+1=0; u=-1/2. Новые вопросы в Алгебра ПОМОГИТЕ ПОЖАЛУЙСТА!!! алгебра 7 класс! фото прикреплю, задание не очень сложное, но я не могу решит) Шахматист играет не менее одной партии в день и не более 12 за любую неделю (неде-ля начинается в понедельник и заканчивается в воскресенье). Докажите … , что найдется несколько подряд идущих дней, за которые он сыграет ровно 20 партий. морская вода содержит 5% солей.Сколько чистой воды надо добавить в 40 литров морской воды,чтобы процент соли был 2%​ найдите числа M и N, чтобы выполнялось равенство (фото ниже) если не тяжело, то с подробным решением Найдите такие числа M и N, чтобы выполнялось равенство(фото ниже) если не тяжело, то с подробным решением ^@[email protected]^ 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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Answer 1

Для первого вопроса, который вы предоставили, я вижу, что вы уже получили ответ на систему уравнений:

1. 2u + 5v = 0 ...............(1)
2. -8u + 15v = 7 ...........(2)

Для системы уравнений вычислили значения переменных: v = 1/5 u = -1/2

Ответ: u = -1/2, v = 1/5

Теперь, перейдем ко второму вопросу:

Задача: Шахматист играет не менее одной партии в день и не более 12 за любую неделю. Докажите, что найдется несколько подряд идущих дней, за которые он сыграет ровно 20 партий.

Для доказательства этого утверждения воспользуемся принципом Дирихле, также известным как принцип ящиков или принцип Дирихле-Штейнера.

Принцип Дирихле: Если n + 1 объектов (шахматные партии в данном случае) размещаются в n ящиках (дни недели), то хотя бы в одном из ящиков будет не менее одного объекта (как минимум одна партия была сыграна в этот день).

Теперь предположим, что шахматист играет ровно 12 партий за каждую неделю. На протяжении всей недели он не может сыграть более 12 партий. Нам нужно добиться, чтобы количество партий было равно 20. Давайте рассмотрим две недели:

1 неделя: 12 партий 2 неделя: 8 партий (20 - 12 = 8)

Теперь у нас есть 20 партий, которые он сыграл в течение 14 дней. В принципе Дирихле обязательно должен существовать хотя бы один день, в котором он сыграл не менее одной партии. Но у нас есть 14 дней, и только 12 партий было сыграно. Если мы добавим 1 партию на каждый из двух дней, которые предшествуют этому дню с одной партией, то мы получим:

1 неделя: 13 партий 2 неделя: 9 партий

Теперь у нас есть 22 партии, которые он сыграл за 16 дней. Принцип Дирихле применяется снова. Мы продолжаем добавлять по одной партии на каждый день, пока не получим:

1 неделя: 16 партий 2 неделя: 4 партии

Теперь у нас есть 20 партий, которые он сыграл за 18 дней. Теперь, если он сыграет еще 1 партию на следующий день (19-й день), у нас будет:

1 неделя: 17 партий 2 неделя: 3 партии

Теперь у нас есть 21 партия, которую он сыграл за 19 дней. И, наконец, если он сыграет еще 1 партию на последующий день (20-й день), у нас будет:

1 неделя: 18 партий 2 неделя: 2 партии

Теперь у нас есть 22 партии, которые он сыграл за 20 дней. Это и есть доказательство того, что найдутся несколько подряд идущих дней, за которые он сыграет ровно 20 партий.