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daniilkomarov
@daniilkomarov
July 2022
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Дамы и господа, помогите решить неравенство!
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sedinalana
Verified answer
ОДЗ (х-4)/(x+4)>0
x=4 x=-4
x∈(-∞;-4) U (4;∞)
(x-4)/(x+4)=a⇒(x+4)/(x-4)=1/a
log(1/3)a-log(1/a)3>0
перейдем на основание 3
log(3)a/log(3)1/3-log(3)3/log(3)1/a=-log(3)a+1/log(3)a>0
(1-log²(3)a)/log(3)a>0
log(3)a=b
(1-b)(1+b)/b>0
b=1 b=-1 b=0
+ _ + _
---------------(-1)------------(0)--------(1)--------------
b<-1 U 0<b<1
1)log(3)a<-1⇒a<1/3
(x-4)/(x+4)<1/3
(x-4)/(x+4)-1/3<0
(3x-12-x-4)/(x+4)<0
(2x-16)/(x+4)<0
x=8 U x=-4
-4<x<8+ОДЗ⇒x∈(4;8)
2)0<log(3)a<1
1<a<3
{(x-4)/(x+4)>1
{(x-4)/(x+4)<3
a)(x-4)/(x+4)-1>0
(x-4-x-4)/(x+4)>
-8/(x+4)>0⇒x+4<0⇒x<-4
b)(x-4)/(x+4)-3<0
(x-4-3x-12)/(x+4)<0
(-2x-16)/(x+4)<0
(2x+16)/(x+4)>0
x=-8 U x=-4
x<-8 U x>-4
нет решения
Ответ x∈(-∞;-4)
1 votes
Thanks 1
daniilkomarov
Большое спасибо, Солнце!
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Answers & Comments
Verified answer
ОДЗ (х-4)/(x+4)>0x=4 x=-4
x∈(-∞;-4) U (4;∞)
(x-4)/(x+4)=a⇒(x+4)/(x-4)=1/a
log(1/3)a-log(1/a)3>0
перейдем на основание 3
log(3)a/log(3)1/3-log(3)3/log(3)1/a=-log(3)a+1/log(3)a>0
(1-log²(3)a)/log(3)a>0
log(3)a=b
(1-b)(1+b)/b>0
b=1 b=-1 b=0
+ _ + _
---------------(-1)------------(0)--------(1)--------------
b<-1 U 0<b<1
1)log(3)a<-1⇒a<1/3
(x-4)/(x+4)<1/3
(x-4)/(x+4)-1/3<0
(3x-12-x-4)/(x+4)<0
(2x-16)/(x+4)<0
x=8 U x=-4
-4<x<8+ОДЗ⇒x∈(4;8)
2)0<log(3)a<1
1<a<3
{(x-4)/(x+4)>1
{(x-4)/(x+4)<3
a)(x-4)/(x+4)-1>0
(x-4-x-4)/(x+4)>
-8/(x+4)>0⇒x+4<0⇒x<-4
b)(x-4)/(x+4)-3<0
(x-4-3x-12)/(x+4)<0
(-2x-16)/(x+4)<0
(2x+16)/(x+4)>0
x=-8 U x=-4
x<-8 U x>-4
нет решения
Ответ x∈(-∞;-4)