Răspuns:
a=571/2415, b=571/1932 si c=571/1380
Explicație pas cu pas: a≠0
0,(1)·(a+b)=0,08(3)·(b+c)=0,0(9)·(c+a) si
144(a+bc)=109(ab+ac)
transformam in fractii primele 3 ecuatii
(a+b)/9=(8/100 +1/300)(b+c)=(a+c)/11
(a+b)/9=(b+c)/12=(a+c)/11 c=11b-12a
(a+b)/9=(b+c)/12 , 4(a+b)=3(b+c) ,
b=3c-4a c=11b-12a=33c-44a-12a 4c=7a, c=7a/4 inlocuim tat
144·[a+3c²-4ac]=109·[3ac-4a²+ac]
144a+3·144·49a²/16 -4·7a²/4= 109·[21a/4-4a²+7a²/4]
144a+1323a²-7a²=2289a/4 - 436a²+763a²/4
576a+5292a²-28a²=2289a -1744a²+763a²
7245a²-1713a=0 [÷3]
2415a²=571a a=571/2415
c=[7×571]/[4×2415]=571/1380
b=3×571/1380 -4×571/2415=571· [ 1/460 - 4/2415]=571/115 ·[1/4 -4/21)=
=571/115 · 5/84= 571/1932
