Question

Cum se rezolva ecuatiile cu partea fractionara si partea intreaga a unui numar?
De ex:

5x+[2x]+[x]+{x}+{2x}+{3x}+{4x}=22

prima parte ar fi sa transformam partile fractionare in "nr - partea intreaga din numar", insa mai departe nu stiu
ms :)

Answer 1

5x+[2x]+[x]+{x}+{2x}+{3x}+{4x}=22 <=>
<=> 5x+[2x]+[x]+x-[x]+2x-[2x]+3x-[3x]+4x-[4x]=22 <=>
<=> 15x - [3x] - [4x] = 22.

3x-1<[3x]≤3x
4x-1<[4x]≤4x
-------------------  (Adunam)
7x-2<[3x]+[4x]≤7x.        (Impartim la (-1))
2-7x>-[3x]-[4x]≥-7x.

Deci 15x-[3x]-[4x]≥15x-7x <=> 22≥8x => x≤$\frac{22}{8}= \frac{11}{4}$

Totodata 15x-[3x]-[4x]<15x+2-7x <=> 22<8x+2 <=> 8x+2>22 <=> 8x>20 => x>$\frac{20}{8}= \frac{5}{2}$.

Deci x∈ $( \frac{5}{2}; \frac{11}{4}]$.

Revenim la 15x-[3x]-[4x]=22. De aici rezulta 15x= 22+[3x]+[4x] . Deci 15x este numar intreg => x∈Z sau x=$\frac{m}{3}$ sau $x= \frac{n}{5}$ sau x=$\frac{p}{15}$. (m,n,p∈Z).

Se obtine solutia $x= \frac{8}{3} .$