Question

vreau raspunsul complet... va ajut si eu daca aveti nevoie.

Answer 1

[3-2,(3)]²:2/3+2,5=1/6xy
2,(3)=2 intregi si3/9=(9x2+3)/9=21/9=7/3

(3-7/3)²:2/3+2,5=xy/6
[(9-7)/3]²:2/3+2,5=xy/6
(2/3)²:2/3+2,5=xy/6
2/3+5/2=xy/6
(4+15)/6=xy/6
19/6=xy/6
xy=19

Fractii echivalente⇒12/(2t+1)=4/5; (6z+4)/35=4/5

12/(2t+1)=4/5
4(2t+1)=12x5
4(2t+1)=60
2t+1=60:4
2t+1=15
2t=15-1
2t=14
t=14:2
t=7

(6z+4)/35=4/5
5(6z+4)=35x4
5(6z+4)=140
6z+4=140:5
6z+4=28
6z=28-4
6z=24
z=24:6
z=4

Deci anul xyzt este:1947