ПОМОГИТЕ РЕШИТЬ НЕРАВЕНСТВА !(ПОДРОБНОЕ РЕШЕНИЕ ). Created by DomnikaMaksimova. algebra-ru

1)x∈(-6; +∞)Ответ: (-6; +∞).2)ОДЗ: 2x-8>0 2x>8 x>4{x>4{x<7x∈(4; 7)Ответ: (4; 7).3)ОДЗ: x-1>0 x>1{x>1{x<5x∈(1; 5)Ответ: (1; 5).4)1) Вариант 1:{x+2>0{0< x-2 < 1{x+2 > x-2a) x+2>0 б) 0< x-2 < 1 в) x+2 > x-2 x> -2 0+2 <x <1+2 x-x > -2-2 2 < x < 3 0 > -4 x - любое число{x>-2{2< x <3{x - любое числоx∈(2; 3)2) Вариант 2:{x+2>0{x-2>1{x+2<x-2a) x+2>0 б) x-2>1 в) x+2<x-2 x> -2 x>1+2 x-x< -2-2 x>3 0 < -4 нет решенийСистема не имеет решений.3) Общее решение неравенства: х∈(2; 3)Ответ: (2; 3).Новые вопросы в Алгебрапомогите пожалуйста решить методом сложения Помогите решить. х=-1 у=5 то 2х+9у-7Помогите решить. х=-1 у=5 то 2х+9у-7Помогите решить. х÷х+9=14ПЛС М+ПОМОГИТЕ, ДАМС КОРОНУ И МНОГО 30 БАЛЛОВ(Нужно объяснение)557,559 номера. Если возможно,с решением Разложить на множетелиy^8-y^4+4y^2-4Дайте відповідь з рішенням будь ласка(557,559)1) 3 1/9 : 2 1/3 - 2 5/6 =28/9 * 3/7 - 17/6=4/3-17/6= 8/6-17/6=-3/2=-1.5как он получил -3/2Іван зірвав на клумбі 8 нарцисів та 4 тюльпани. Скільки всього існуєспособів вибору із цих квітів 3 нарцисів та 2 тюльпанів для букета?ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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ПОМОГИТЕ РЕШИТЬ НЕРАВЕНСТВА !(ПОДРОБНОЕ РЕШЕНИЕ )...

DomnikaMaksimova @DomnikaMaksimova

August 2022 1 28 Report

ПОМОГИТЕ РЕШИТЬ НЕРАВЕНСТВА !(ПОДРОБНОЕ РЕШЕНИЕ )

Answers & Comments

m11m

Verified answer

1)
$lg3^{x-1}-lg3^{2x+4}\ \textless \ lg3 \\ \\ 
lg( \frac{3^{x-1}}{3^{2x+4}} )\ \textless \ lg3 \\ \\ 
lg(3^{x-1-(2x+4)})\ \textless \ lg3 \\ \\ 
lg(3^{x-1-2x-4})\ \textless \ lg3 \\ \\ 
lg(3^{-x-5})\ \textless \ lg3 \\ \\ 
3^{-x-5}\ \textless \ 3^1 \\ \\ 
-x-5\ \textless \ 1 \\ 
-x\ \textless \ 1+5 \\ 
-x\ \textless \ 6 \\ 
x\ \textgreater \ -6$

x∈(-6; +∞)
Ответ: (-6; +∞).

2)
ОДЗ: 2x-8>0
2x>8
x>4

$log_{3.1}(2x-8)-log_{3.1}6\ \textless \ 0 \\ \\ 
log_{3.1}( \frac{2x-8}{6} )\ \textless \ 0 \\ \\ 
 \frac{2x-8}{6}\ \textless \ 3.1^0 \\ \\ 
 \frac{2x-8}{6}\ \textless \ 1 \\ \\ 
2x-8\ \textless \ 6 \\ 
2x\ \textless \ 6+8 \\ 
2x\ \textless \ 14 \\ 
x\ \textless \ 7$

{x>4
{x<7

x∈(4; 7)
Ответ: (4; 7).

3)
ОДЗ: x-1>0
x>1

$log_{ \frac{1}{ \sqrt{2} } }(x-1)+log_{2}(x-1)\ \textgreater \ -2 \\ \\ 
log_{( \sqrt{2} )^{-1}}(x-1)+log_{( \sqrt{2} )^2}(x-1)\ \textgreater \ -2 \\ \\ 
-log_{ \sqrt{2} }(x-1)+ \frac{1}{2}log_{ \sqrt{2} }(x-1)\ \textgreater \ -2 \\ \\ 
- \frac{1}{2}log_{ \sqrt{2} }(x-1)\ \textgreater \ -2 \\ \\ 
log_{ \sqrt{2} }(x-1)\ \textless \ 4 \\ \\ 
x-1\ \textless \ ( \sqrt{2} )^4 \\ \\ 
x-1\ \textless \ 4 \\ 
x\ \textless \ 4+1 \\ 
x\ \textless \ 5$

{x>1
{x<5

x∈(1; 5)
Ответ: (1; 5).

4)
1) Вариант 1:
{x+2>0
{0< x-2 < 1
{x+2 > x-2

a) x+2>0 б) 0< x-2 < 1 в) x+2 > x-2
x> -2 0+2 <x <1+2 x-x > -2-2
2 < x < 3 0 > -4
x - любое число
{x>-2
{2< x <3
{x - любое число

x∈(2; 3)

2) Вариант 2:
{x+2>0
{x-2>1
{x+2<x-2

a) x+2>0 б) x-2>1 в) x+2<x-2
x> -2 x>1+2 x-x< -2-2
x>3 0 < -4
нет решений
Система не имеет решений.

3) Общее решение неравенства:
х∈(2; 3)

Ответ: (2; 3).