nє(-∞;1/3]U[3;+∞)
Объяснение:
2sin(x)=(n+1)/(n-1)sin(x)=(n+1)/2(n-1)sin(x)=(n+1)/(2n-2)-1≤sin(x)≤1-1≤(n+1)/(2n-2)≤11)-1≤(n+1)/(2n-2)(n+1)/(2n-2)≥-1
(n+1)/(2n-2) + 1≥0
(n+1)/(2n-2) + (2n-2)/(2n-2) ≥0
(n+1+2n-2)/(2n-2)≥0
(3n-1)/(2n-2)≥0
ОДЗ: 2n-2≠0
(2n≠2,n≠1)
(3n-1)*(2n-2)≥0 (n≠1)
(n-1/3)*(n-1)≥0 (n≠1)
+. —. +.
•o›
1/3. 1. n
nє(-∞;1/3]U(1;+∞)
2)(n+1)/(2n-2)≤1(n+1)/(2n-2)-1≤0
(n+1)/(2n-2)-(2n-2)/(2n-2)≤0
(n+1-2n+2)/(2n-2)≤0
(-n+3)/(2n-2)≤0
(n-3)/(2n-2)≥0
ОДЗ:2n-2≠0
2n≠2, n≠1
(n-3)*(2n-2)≥0 (n≠1)
(n-3)*(n-1)≥0. (n≠1)
+. —. +.
o•›
1. 3. n
nє(-∞;1)U[3;+∞)
объединяем:nє(-∞;1/3]U(1;+∞) и nє(-∞;1)U[3;+∞)
1/3. 1. 3
•о•›
. n
значит nє(-∞;1/3]U[3;+∞)
