Question

Va rog frumos sa mi rezolvati si mie exercitiul 5 acela incercuit de mine!!

Answer 1

rezolvam ecuatia scotand mai intai {a}=partea fractionara a lui a
 si
[a]=partea intreaga din a
prin calcul direct ecuatia devine
x+2*1+3 *1/6=2-[2-13/3] +2/3+2*(2-√2)

Am tinut cont ca ;

[5/3]=[1 si 2/3] =1

0<0.1(6)<1 deci {0,1(6)}=0,1(6)=1/6
4 , (3)= 4 intregi si 1/3=13/3

{2/3}=2/3 pt ca 0<2/3<1

-2<-√2<-1
{√2}=-√2-[-√2]=-√2-(-2)=-√2+2=2-√2

deci
x+2*1+3 *1/6=2-[2-13/3] +2/3+2*(2-√2)
x+2+1/2=2-[6/3-13/3} +2/3+2(2-√2)

[6/3-13/3]=[-7/3]=[-2si1/3]=-3
asadar ecuatia devine
x+5/2=2-(-3) +2/3+4-2√2
x+5/2=5+4+2/3-2√2
x=5/2+4+2/3 -2√2

x= (15+24+4)/6-2√2
x=43/6-2√2
x=7si1/6-2√2

aproximam  x cu o aproximatie suficienta, cu un numar rational,  pt a ii afla partea intreaga
x≈4,338
deci [x]=4
atunci
{x}=x-[x]= 43/6-2√2-4=43/6-24/6-2√2=19/6-2√2
[x]=4
{x}=19/5-2√2