[ x+1 ] = 4
4 <= x + 1 < 5
3 <= x < 4
x € [3 , 4)
[3x - 6 ] = 7
7<= 3x - 6 < 8
13 <= 3x < 14
13/3 <= x < 14/3
x€ [13/3 , 14/3)
[ x ] + [ x] + [x] + 1 + 2 = x + 5/2
3[x] = x - 1/2
[x] = (x -1/2)/3
[x] = (2x-1)/6
(2x-1)/6 = k -> 2x - 1 = 6k , 2x = 6k + 1 , x = (6k+1)/2
k <= (6k+1)/2 < k + 1
2k <= 6k+1 < 2k +2
2k - 1 <= 6k < 2k + 1
2k - 1 <= 6k
-4k <= 1 -> k >= -1/4 , k€ [-1/4 , + infinit)
6k + 1 < 2k + 2
4k < 1 -> k € ( - infinit , 1/4)
k€ (- 1/4 , 1/4]
k = 0 -> x = 1/2
d) [x] + [x-3] = 1 + [x+4]
[x] + [x] - 3 = 1 + [x] + 4
[x] = 8
8 <= x < 9 -> x € [8,9)
e)[x] = 2 - x = k
x = 2 - k
k<= 2 - k < k + 1
k <= 2 - k
2k <= 2 , k <= 1 , k€( - infinit , 1]
2 - k < k + 1
-2k < 1/2 , k > -1/4 , k € ( -1/4 , + infinit)
k€ { 0 , 1}
x = 2
x = 1
pe f il rezolvi si tu.
