Предмет: Математика, автор: vd331783

Решите неравенство log3(x²-2x+1)<=2


a87086828: log3(x²-2x+1)-log33<=0
log3(x²-2x+1)-1<=0
log3(x²-2x+1)<=1 Теперь можем применить свойства логарифмов:
x²-2x+1<=3
(x-1)²<=3
x²-2x+1<=3
x²-2x<=2
x²-2x-2<=0
(x-4)(x+1)<=0
Итак, решение неравенства log3(x²-2x+1)<=2 - это промежуток (-∞;-1) U (4;+∞).
Ответ:
(-∞;-1) U (4;+∞)
vd331783: А можете с D
a87086828: я не понял
vd331783: С дискриминантом решить это же
a87086828: а хорошо
a87086828: можете удалить мое задание чтобы я переписал дискриминантом?

Ответы

Автор ответа: a87086828
0

Ответ:

Пошаговое объяснение:

Сначала найдем область допустимых значений (ОДЗ) для неравенства. Для этого рассмотрим функцию log3(x²-2x+1) и определим ее область определения.

Функция log3(x²-2x+1) определена для всех x, таких что x²-2x+1>0. Решая это квадратное уравнение, получаем x²-2x+1=(x-1)²>0, что верно для всех x≠1. Таким образом, ОДЗ для неравенства log3(x²-2x+1)<=2 - это промежуток (-∞;1) U (1;+∞).

Теперь, когда мы знаем ОДЗ, можем приступить к решению неравенства.

Из неравенства log3(x²-2x+1)<=2 получаем, что log3(x²-2x+1)-2<=0.

По правилу разности логарифмов, получаем:

log3(x²-2x+1)-log33<=0

log3(x²-2x+1)-1<=0

log3(x²-2x+1)<=1

Теперь можем применить свойства логарифмов:

x²-2x+1<=3

(x-1)²<=3

x²-2x+1<=3

x²-2x<=2

x²-2x-2<=0

(x-4)(x+1)<=0

Итак, решение неравенства log3(x²-2x+1)<=2 - это промежуток (-∞;-1) U (4;+∞).

Ответ:

(-∞;-1) U (4;+∞)

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