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workinglogorifm
@workinglogorifm
August 2021
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решите пример, 100 баллов...
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oganesbagoyan
Verified answer
Log_1/7 (5x+3) ≥ -1/2 ⇔0 < 5x+3 ≤ (1/7)^(-1/2) .
т.к. 0 < 1/7 < 1.
---
0 < 5x+3
≤ ((7)^(-1) ) ^(-1/2) ;
0 < 5x+3 ≤ √
7;
-3 <5x ≤ √
7 -3 ;
-3/5 <x ≤ (√
7 -3)/5.
ответ0: x ∈(-3/5 ; (√
7 -3)/5 ] .
1 votes
Thanks 1
oganesbagoyan
ответ: x ∈(-3/5 ; (-3+√7)/5 ] .
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Answers & Comments
Verified answer
Log_1/7 (5x+3) ≥ -1/2 ⇔0 < 5x+3 ≤ (1/7)^(-1/2) .т.к. 0 < 1/7 < 1.
---
0 < 5x+3 ≤ ((7)^(-1) ) ^(-1/2) ;
0 < 5x+3 ≤ √7;
-3 <5x ≤ √7 -3 ;
-3/5 <x ≤ (√7 -3)/5.
ответ0: x ∈(-3/5 ; (√7 -3)/5 ] .