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micklson
@micklson
April 2021
2
22
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Решите логарифмические неравенства
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sedinalana
Verified answer
1
ОДЗ
x+2>0⇒x>-2
x>0
2x+1>0⇒x>-0,5
x∈(0;∞)
log(3)[x(x+2)]<log(3)(2x+1)
x²+2x<2x+1
x²-1<0
(x-1)(x+1)<0
x=1 x=-4
-1<x<1
x⇒(0;1)
2
ОДЗ
49-x²≥0
(7-x)(7+x)≥0
x=7 x=-7
-7≤x≤7
x>0
x∈(0;7]
√(49-x²)≥0⇒log(5)x/(x-5)≥0
1)log(5)x≥0⇒x≥1
x-5>0⇒x>5
5<x≤7
2)log(5)x≤0⇒x≤1
x-5<0⇒x<5
0<x≤1
x∈(0;1] U (5;7]
1 votes
Thanks 1
m11m
Verified answer
1)
ОДЗ: 1) x+2>0 2) x>0 3) 2x+1>0
x> -2 2x> -1
x> -0.5
{x>-2
{x>0
{x> -0.5
В итоге x>0
x(x+2)<2x+1
x²+2x-2x-1<0
x²-1<0
(x-1)(x+1)<0
x=1 x= -1
+ - +
------- -1
------------
1 -----------
\\\\\\\\\\\\\\
x∈(-1; 1)
C учетом ОДЗ:
{x>0
{x∈(-1; 1)
x∈(0; 1)
Ответ: (0; 1).
2)
ОДЗ: 1) 49-x²≥0 2) x>0 3) x-5≠0
x²-49≤0 x≠5
(x-7)(x+7)≤0
x=7 x= -7
+ - +
-------
-7 -------- 7
-----------
\\\\\\\\\\\
x∈[-7; 7]
{x∈[-7; 7]
{x>0
{x≠5
В итоге ОДЗ: x∈(0; 5)U(5; 7].
√(49-x²)=0 log₅ x=0 x-5=0
x₁=7 x=1 x=5
x₂= -7(не под-
ходит по ОДЗ)
+ - +
0
---------- 1
---------- 5
------------ 7
\\\\\\\\\\\\ \\\\\\\\\\\\\\\
х=6 √(49-6²)>0 (+)
log₅ 6>0 (+) ⇒ (+)
6-5>0 (+)
x=2 √(49-2²)>0 (+)
log₅ 2>0 (+) ⇒ (-)
2-5<0 (-)
x=0.5 √(49-0.5²)>0 (+)
log₅ 0.5<0 (-) ⇒ (+)
0.5-5<0 (-)
x∈(0; 1]U(5; 7]
Ответ: (0; 1]U(5; 7].
0 votes
Thanks 0
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Answers & Comments
Verified answer
1ОДЗ
x+2>0⇒x>-2
x>0
2x+1>0⇒x>-0,5
x∈(0;∞)
log(3)[x(x+2)]<log(3)(2x+1)
x²+2x<2x+1
x²-1<0
(x-1)(x+1)<0
x=1 x=-4
-1<x<1
x⇒(0;1)
2
ОДЗ
49-x²≥0
(7-x)(7+x)≥0
x=7 x=-7
-7≤x≤7
x>0
x∈(0;7]
√(49-x²)≥0⇒log(5)x/(x-5)≥0
1)log(5)x≥0⇒x≥1
x-5>0⇒x>5
5<x≤7
2)log(5)x≤0⇒x≤1
x-5<0⇒x<5
0<x≤1
x∈(0;1] U (5;7]
Verified answer
1)ОДЗ: 1) x+2>0 2) x>0 3) 2x+1>0
x> -2 2x> -1
x> -0.5
{x>-2
{x>0
{x> -0.5
В итоге x>0
x(x+2)<2x+1
x²+2x-2x-1<0
x²-1<0
(x-1)(x+1)<0
x=1 x= -1
+ - +
------- -1 ------------ 1 -----------
\\\\\\\\\\\\\\
x∈(-1; 1)
C учетом ОДЗ:
{x>0
{x∈(-1; 1)
x∈(0; 1)
Ответ: (0; 1).
2)
ОДЗ: 1) 49-x²≥0 2) x>0 3) x-5≠0
x²-49≤0 x≠5
(x-7)(x+7)≤0
x=7 x= -7
+ - +
------- -7 -------- 7 -----------
\\\\\\\\\\\
x∈[-7; 7]
{x∈[-7; 7]
{x>0
{x≠5
В итоге ОДЗ: x∈(0; 5)U(5; 7].
√(49-x²)=0 log₅ x=0 x-5=0
x₁=7 x=1 x=5
x₂= -7(не под-
ходит по ОДЗ)
+ - +
0 ---------- 1 ---------- 5 ------------ 7
\\\\\\\\\\\\ \\\\\\\\\\\\\\\
х=6 √(49-6²)>0 (+)
log₅ 6>0 (+) ⇒ (+)
6-5>0 (+)
x=2 √(49-2²)>0 (+)
log₅ 2>0 (+) ⇒ (-)
2-5<0 (-)
x=0.5 √(49-0.5²)>0 (+)
log₅ 0.5<0 (-) ⇒ (+)
0.5-5<0 (-)
x∈(0; 1]U(5; 7]
Ответ: (0; 1]U(5; 7].