натуральные числа, разность квадратов которых равна числу 2017. Created by vine3. algebra-ru

A²-b²=2017(a-b)·(a+b)=2017Разложим 2017 на множители.2017- число простое2017=1·2017Значит {a-b=1{a+b=2017Cкладываем2a=2018a=1009b=1008О т в е т. 1008 и 1009 эм, а какие собственно числа?Новые вопросы в Алгебрапомогите с 3,4,5. дам много баловX sin x cos x tg x ctg x π+a - sin a - cos a tg a ctg a π-A sin a cos a - tg a - ctg a 2πτα sin a cos a tg a ctg a 2π-a - sin a cos a - tg a - ctg решите задачу пожалуйстадокажите что для любого натурального числа n справедливо неравенство :1/2^2 + 1/4^2 + 1/6^2 + 1/(2n)^2 + ... < 1/2решите задачу пожалуйстаРешите пожалуйста задачуРешите задачу пожалуйстаРешите задачу пожалуйстаякого найменшого значення набуває вираз і якого значення х х^2-10х+26Решиет уравнение :x^2 + y^2 = 4x^3 + y^3 = -8ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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натуральные числа, разность квадратов которых равна числу 2017...

vine3 @vine3

July 2022 1 1 Report

натуральные числа, разность квадратов которых равна числу 2017

Answers & Comments

nafanya2014

Verified answer

A²-b²=2017
(a-b)·(a+b)=2017
Разложим 2017 на множители.
2017- число простое
2017=1·2017
Значит
{a-b=1
{a+b=2017
Cкладываем
2a=2018
a=1009
b=1008
О т в е т. 1008 и 1009