Question

DAU COROANĂ!!! ESTE URGENT!!! Exercițiul 9, vă rog mult de tot!

Answer 1

a) [x]=2 <=> 2<=x<3 deci x€[2;3)

b) [3x]=1 <=> 1/3<=x<2/3 deci x€[1/3;2/3)

c) {2x}=0 <=> 2x € Z <=>

x €{...-1/2,0,1/2...}

d) {x}=3/4

x€{x>0| x=y+3/4, y€N} U {x<0| x=-y-1/4,y€N}

e) (x-1)/2<=(3x-1)/4<(x+1)/2

2x-2<=3x-1<2x+2

-1<= x< 3

x€[-1;3)

f) x-3<= x/2 < x-2

2x-6<=x< 2x-4 |-2x

-6<= -x < -4 |*(-1)

6>= x> 4

x€(4;6]

Answer 2

a.  [x]=2 => 2 <= x < 3 dar x∈R <=> x∈[2,3) .

b.  [3x]=1 => 1 <= 3x <2 <=> 1/3 <= x < 2/3 dar x∈R <=> x∈[1/3,2/3) .

c.  {2x}=0 din 2x=[2x]+{2x} pentru orice x∈R => [2x]=2x dar [2x]∈Z <=> 2x∈Z <=> x∈Z ;

d.  {x}=3/4 => x∈R₊ a.i. sa indeplineasca proprietatea din stanga ;

e.  [3x-1 /4]=x-1 /2 => [3x-1 /4]=k ,unde k∈Z => x=2k+1 => k <=3x-1 /4 < k+1 <=> k <= 6k+2 /4 < k+1 => 4k <= 6k+2 => 0 <= 2k+2 <=> 0 <= k+1; si 6k+2 < 4k+4 => 2k+2 < 4 => k+1 < 2 => 0 <= k+1 < 2 iar singura varianta posibila pentru k∈Z este k=0 => x=1 .

f.  [x/2]=x-3 => [x/2]=k ,unde k∈Z => x=k+3 => k <= x/2 < k+1 => k <= k+3 /2 <k+1 => 2k <=k+3 => k <= 3 si k+3 < 2k+2 => 3 < k+2 => 1 < k => 1 < k <=  3 => k∈{2;3} => x∈{5;6} .