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aniyn
@aniyn
July 2022
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Решить три примера
Срочно , до завтра . Кто что может
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ВладимирБ
Verified answer
Решение смотри на фото
0 votes
Thanks 1
oganesbagoyan
Verified answer
Task/25353447
-------------------- см. приложение
3.
решите неравенство Log(2) (x² +2x+1) + Log(0,5) (x+1) ≤ 0 .
-------
Log(2) (x² +2x+1) +
Log(2⁻¹ ) (x+1) ≤ 0 ;
2Log(2) (x+1)
-
Log(2) (x+1) ≤ 0 ; * * * ОДЗ: x+1 >0 * * *
Log(2) (x+1) ≤
Log(2) 1
;
0< x+1 ≤1⇔ -1 < x ≤ 0 .
ответ :
x
∈ ( -1 ;
0 ] .
-------
4.
Решите уравнение 2(sinx - cosx) = tgx -1.
-------
2(sinx - cosx) =sinx / cosx -1 ; * * * ОДЗ: cosx ≠ 0 ⇒x ≠π/2 +πn , n ∈ Z * * *
2(sinx - cosx)cosx =(sinx - cosx) ;
2(
sinx - cosx)(cosx -1/2) =0 ;
а)
sinx -cosx = 0 ; * * * √2sin(x-π/4) =0⇒ x - π/4 = πn * * *
sinx =cosx || : cosx ≠0
tgx =1 ;
x= π/4 +πn ,n ∈ Z.
б)
cosx -1/2 = 0 ;
cosx =1/2 ;
x =± π/3 +2πk , k∈Z.
ответ :
π/4 +πn , n ∈ Z
±
π/3 +2πk , k∈Z
.
0 votes
Thanks 0
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Answers & Comments
Verified answer
Решение смотри на фотоVerified answer
Task/25353447-------------------- см. приложение
3.
решите неравенство Log(2) (x² +2x+1) + Log(0,5) (x+1) ≤ 0 .
-------
Log(2) (x² +2x+1) + Log(2⁻¹ ) (x+1) ≤ 0 ;
2Log(2) (x+1) - Log(2) (x+1) ≤ 0 ; * * * ОДЗ: x+1 >0 * * *
Log(2) (x+1) ≤ Log(2) 1 ;
0< x+1 ≤1⇔ -1 < x ≤ 0 .
ответ : x ∈ ( -1 ; 0 ] .
-------
4.
Решите уравнение 2(sinx - cosx) = tgx -1.
-------
2(sinx - cosx) =sinx / cosx -1 ; * * * ОДЗ: cosx ≠ 0 ⇒x ≠π/2 +πn , n ∈ Z * * *
2(sinx - cosx)cosx =(sinx - cosx) ;
2(sinx - cosx)(cosx -1/2) =0 ;
а)
sinx -cosx = 0 ; * * * √2sin(x-π/4) =0⇒ x - π/4 = πn * * *
sinx =cosx || : cosx ≠0
tgx =1 ;
x= π/4 +πn ,n ∈ Z.
б)
cosx -1/2 = 0 ;
cosx =1/2 ;
x =± π/3 +2πk , k∈Z.
ответ : π/4 +πn , n ∈ Z
± π/3 +2πk , k∈Z .