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Alichik
@Alichik
April 2021
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Для Sedinalana с полными ответами пожалуйста
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sedinalana
Verified answer
6
ОДЗ
x+3>0⇒x>-3
x+15>0⇒x>-15
x∈(-3;∞)
log(a)x+log(a)y=log(a)(x*y)⇒
log(4)[(x+3)(x+15)]≤3
(x+3)(x+15)≤64
x²+15x+3x+45-64≤0
x²+18x-19≤0
x1+x2=-18 U x1*x2=-19
x1=-19 U x2=1
+ _ +
------------[-19]------------------[1]------------------
-19≤x≤1 +ОДЗ
x∈(-3;1]
7
4cos^4x-4cos²x+1=0
a²-2ab+b²=(a-b)²⇒
(2cos²x-1)²=1
2cos²x-1=-1 U 2cos²x-1=1
1)2cos²x=0
cosx=0
x=π/2+πk,k∈z
-2π≤π/2+πk≤-π
-4≤1+2k≤-2
-5≤2k≤-3
-2,5≤k≤-1,5
k=-2⇒x=π/2-2π=-3π/2
2)2cos²x=2
cos²x=1
(1+cos2x)/2=1
1+cos2x=2
cos2x=1
2x=2πn
x=πn,n∈z
-2π≤πn≤-π
-2≤n≤-1
n=-2⇒x=-2π
n=-1⇒x=-π
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Answers & Comments
Verified answer
6ОДЗ
x+3>0⇒x>-3
x+15>0⇒x>-15
x∈(-3;∞)
log(a)x+log(a)y=log(a)(x*y)⇒
log(4)[(x+3)(x+15)]≤3
(x+3)(x+15)≤64
x²+15x+3x+45-64≤0
x²+18x-19≤0
x1+x2=-18 U x1*x2=-19
x1=-19 U x2=1
+ _ +
------------[-19]------------------[1]------------------
-19≤x≤1 +ОДЗ
x∈(-3;1]
7
4cos^4x-4cos²x+1=0
a²-2ab+b²=(a-b)²⇒
(2cos²x-1)²=1
2cos²x-1=-1 U 2cos²x-1=1
1)2cos²x=0
cosx=0
x=π/2+πk,k∈z
-2π≤π/2+πk≤-π
-4≤1+2k≤-2
-5≤2k≤-3
-2,5≤k≤-1,5
k=-2⇒x=π/2-2π=-3π/2
2)2cos²x=2
cos²x=1
(1+cos2x)/2=1
1+cos2x=2
cos2x=1
2x=2πn
x=πn,n∈z
-2π≤πn≤-π
-2≤n≤-1
n=-2⇒x=-2π
n=-1⇒x=-π