Răspuns :
4 intregi 1/5 + (X-5 intregi 4/5)=2intregi 1/2
(4×5+1)/5 +[X-(5×5+4)/5]=(2×2+1)/2
21/5+(X- 29/5)= 5/2
21/5 + (5X - 29)/5 = 5/2
(21+5X-29)/5= 5/2
(5X - 8)/5=5/2
2(5X-8) = 5×5
10X- 16 = 25
10X = 25+16
10X = 41
X = 41 /10 (apartine multimii numerelor rationale )
(4×5+1)/5 +[X-(5×5+4)/5]=(2×2+1)/2
21/5+(X- 29/5)= 5/2
21/5 + (5X - 29)/5 = 5/2
(21+5X-29)/5= 5/2
(5X - 8)/5=5/2
2(5X-8) = 5×5
10X- 16 = 25
10X = 25+16
10X = 41
X = 41 /10 (apartine multimii numerelor rationale )