Question

daca are cineva timp sa se uite pe exercitiu ...nu ajung la raspunsul din carte...multumesc!!​

Answer 1

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