[(x+3)/2]∈Z
deci
(4x+5) divizibil cu 3
dar
[x]≤x<[x]+1
atunci
(4x+5)/3≤(x+3)/2<(4x+8)/3 amplificam cu 6
8x+10≤3x+9<8x+16 |scadem 8x
10≤9-5x<16 | inmultim cu -1
-10≥5x-9>-16 | +9
-1≥5x>-7 |:5
-0,2≥x>-1,2
x∈(-1,2;-0,2] si 4x+5 divizibil cu 3 adica ∈(...-3;0;3;6;9...}
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4x+5=-3 4x=-8...x=-2 nu convine
4x+5=0 4x=-5 x=-1,25 nu convine
4x+5=3 4x=-2 x=-0,5 convine
4x+5=6 4x=-1 x=-0,25 convine
4x+5=9 4x=4 x=1 nu convine
.....................................................
∑|x| =|-1,25|+|-0,5|=1,25+0,5=1,75=7/4
grea rau!!!
