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Task/2592065-------------------2) {2x³ +x² +ax +b = (x+1)q₁(x) +18 ; { 2(-1)³ +(-1)² +a*(-1) +b=18 ;{2x³ +x² +ax +b =(x-2)q₂. { 2*2³ +2² +2a +b = 0.------{ -a + b = 19 ; { 3a = -20 -19 ; {a = -13{ 2a +b = -20 . { b = -20 -2a . { b =6.ответ: -13 ; 6.====================3) Доказать, что x³ +5 не делится на приведенный квадратный трехчлен сцелыми коэффициентами .---Допустим можно :x³ +5 = (x² +px +q) (x-x₀) , где p, q ∈Z (целые числа) .x₀³ +5 =(x²₀ +px₀ +q) (x₀-x₀) =0 ⇒ x₀ = -∛5 , но 5 = - q*x₀ ⇔ 5 = q* ∛5 _не выполняется при q ∈ Z : получили противоречие , т.е (x³ +5) не делится на приведенный квадратный трехчлен с целыми коэффициентами Реклама Реклама Новые вопросы в Алгебра Решить 5) с графиком и решением 6,7,9 Докину 50 балов 2,39секунды перевести в метры помогите срочно Реши неравенство 2a3<3.(В первое окошко введи знак «<» или «>», во второе окошко введи число). пожалуйста помогите Срочно!!! кто из Домашней школы Интернет урок, киньте ссылку в группу вк с ответами. Никак не могу найти упростите выревжегия Решите пж пж буду благодарен Розв'яжіть нерівність 11 - 2x < 15 – 4х. чи є розв'язками цієї нерівності числа -3; 2? СРОЧНО!!! ДАЮ 100 БАЛЛОВ!!!Пожалуйста, помогите представить в виде корня, задачи с фото!! С полным решением!!! СРОЧНО!!! ДАЮ 100 БАЛЛОВ!!!Пожалуйста, помогите определить четные и нечётные функции. С полным решением!!! Задания на на фотографии!!! Предыдущий Следующий Реклама - это не так уж плохо Реклама - это не так уж плохо Понимаем - вы пришли сюда не за рекламой. Но она помогает нам предоставлять бесплатный доступ к Знаниям. Пожалуйста, внесите наш сайт в список разрешенных. ОК, разблокирую рекламу на Знаниях! 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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Помогите, пожалуйста, очень нужно ❤️...

TequilaSunrise @TequilaSunrise

September 2021 1 70 Report

Помогите, пожалуйста, очень нужно ❤️

Answers & Comments

oganesbagoyan

Verified answer

Task/2592065
-------------------
2)
{2x³ +x² +ax +b = (x+1)q₁(x) +18 ; { 2(-1)³ +(-1)² +a*(-1) +b=18 ;
{2x³ +x² +ax +b =(x-2)q₂.   { 2*2³ +2² +2a +b = 0.
------
{ -a + b = 19 ; { 3a = -20 -19 ; {a = -13
{ 2a +b = -20 . { b = -20 -2a . { b =6.

ответ: -13 ; 6.
====================
3) Доказать, что x³ +5 не делится на приведенный квадратный трехчлен с
целыми коэффициентами .
---
Допустим можно :
x³ +5 = (x² +px +q) (x-x₀) , где  p, q ∈Z (целые числа) .
x₀³ +5 =(x²₀ +px₀ +q) (x₀-x₀) =0 ⇒ x₀ = -∛5 , но 5 = - q*x₀ ⇔  5 = q* ∛5 _не выполняется при q ∈ Z : получили противоречие , т.е (x³ +5)  не делится на приведенный квадратный трехчлен с целыми коэффициентами