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2.Розв'яжіть нерівність: {5-х<2 {12+x<7 Срочно !!! Смотреть ответ abdurokhmannamhkorud abdurokhmannamhkorud Ответ: {18=х8......................... Новые вопросы в Алгебра При каком значении x выражение |6-10|-|-3+(-1)|+x равно нулю? Найдите значение выражения 9!: 3!4! помогите пожалуйста Найдите сумму первых 20 членов арифметической прогрессии, если известно, что a5 + a6 = 28; a8 + a9 = 58.​ 4. Туюнтманын маанисин тапкыла: Найдите значение выражения: 9! 3!4! A) 240 Б) 280 B) 560 Г) 2520​ 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

Давайте по очереди рассмотрим каждое из ваших заданий.

2. Решение неравенства: Неравенство имеет два компонента: 1. \(5 - x < 2\) 2. \(12 + x < 7\)

Для первого компонента: \[5 - x < 2\]

Вычитаем 5 из обеих сторон: \[-x < 2 - 5\] \[-x < -3\]

Меняем знак на противоположный и учтем, что \(a < b\) влечет за собой \(-a > -b\): \[x > 3\]

Для второго компонента: \[12 + x < 7\]

Вычитаем 12 из обеих сторон: \[x < 7 - 12\] \[x < -5\]

Теперь у нас есть два неравенства: 1. \(x > 3\) 2. \(x < -5\)

Первое неравенство означает, что \(x\) должно быть больше 3, а второе неравенство означает, что \(x\) должно быть меньше -5. Поскольку нет значения \(x\), которое одновременно больше 3 и меньше -5, то это неравенство не имеет решений.

По остальным заданиям:

3. При каком значении \(x\) выражение \(|6-10|-|-3+(-1)|+x\) равно нулю? \[|6-10|-|-3+(-1)|+x\]

Вычислим значения внутри модулей: \[|(-4)|-|-3-1|+x\]

\[4-|(-4)|+x\]

\[4-4+x\]

\[0+x\]

\[x\]

Значит, данное выражение равно нулю, когда \(x = 0\).

4. Найдите значение выражения \(9! : (3! \cdot 4!)\): Сначала вычислим факториалы: \(3! = 3 \cdot 2 \cdot 1 = 6\) \(4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24\) \(9! = 9 \cdot 8 \cdot 7 \cdot 6! = 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 362880\)

Теперь вычислим значение выражения: \(9! : (3! \cdot 4!) = 362880 : (6 \cdot 24) = 362880 : 144 = 2520\)

Ответ: \(2520\).

5. Найдите сумму первых 20 членов арифметической прогрессии, если известно, что \(a_5 + a_6 = 28\) и \(a_8 + a_9 = 58\).

Общий вид члена арифметической прогрессии: \(a_n = a_1 + (n-1)d\), где \(a_n\) - n-й член прогрессии, \(a_1\) - первый член прогрессии, \(n\) - номер члена, \(d\) - разность между соседними членами.

Сначала найдем разность \(d\): \[a_6 - a_5 = 28 - a_5\] \[a_9 - a_8 = 58 - a_8\]

Так как это арифметическая прогрессия, то разность между соседними членами одинакова: \[28 - a_5 = 58 - a_8\] \[a_8 - a_5 = 58 - 28\] \[a_8 - a_5 = 30\]

Теперь мы знаем разность \(d\).

Теперь найдем первый член \(a_1\): \[a_6 = a_1 + (6-1)d\] \[a_5 = a_1 + (5-1)d\]

Мы знаем, что \(a_6 - a_5 = 28\), поэтому подставим значения: \[a_1 + (6-1)d - (a_1 + (5-1)d) = 28\] \[a_1 + 5d - a_1 - 4d = 28\] \[d = 28\]

Теперь у нас есть значение разности \(d\) и значение \(a_8 - a_5 = 30\).

Теперь можно найти сумму первых 20 членов: \[S_{20} = \frac{n}{2}(2a_1 + (n-1)d)\] \[S_{20} = \frac{20}{2}(2a_1 + (20-1) \cdot 28)\] \[S_{20} = 10(2a_1 + 19 \cdot 28)\] \[S_{20} = 10(2a_1 + 532)\] \[S_{20} = 10(2(a_1 + 532))\]

Теперь нам нужно найти значение \(a_1\), которое мы можем найти, используя разность \(d\) и одно из уравнений: \[a_1 + 5d = 28\] \[a_1 + 5 \cdot 28 = 28\] \[a_1 + 140 = 28\] \[a_1 = 28 - 140\] \[a_1 = -112\]

Теперь подставим значение \(a_1\) в выражение для суммы: \[S_{20}