Question

aici e poza la intrebarea anterioara

Answer 1

$\displaystyle\it\\\frac{2a}{2+a}=\frac{3b}{3+b}=\frac{4c}{4+c} \implies\\\frac{2a}{2+a}=\frac{3b}{3+b} \Leftrightarrow 2a(3+b)=3b(2+a) \Leftrightarrow 6a+2ab=6b+3ab \Leftrightarrow\\ab-6a+6b=0 |-36 \implies (a+6)(b-6)=-36.\\\frac{3b}{3+b}=\frac{4c}{4+c} \Leftrightarrow 3b(4+c)=4c(3+b) \Leftrightarrow 12b+3bc=12c+4bc \Leftrightarrow\\bc+12c-12b=0 |-144 \implies (b+12)(c-12)=-144.\\\frac{2a}{2+a}=\frac{4c}{4+c} \Leftrightarrow 2a(4+c)=4c(2+a) \Leftrightarrow 8a+2ac=8c+4ac |:2$

$\displaystyle\it\\\implies 4a+ac=4c+2ac \Leftrightarrow 4c+ac-4a=0|-16\\ \implies (4+a)(c-4)=-16,~pentru~ca~(4+a)(c-4)~este~negativ,~iar~\\a,c\in\mathbb{N} \implies c-4 = negativ \implies c\in\left\{1,2,3\right\}.\\daca~c=1 \implies -3(4+a)=-16,~dar~3~nu~divide~pe~16,\\acest~caz~cade.\\daca~c=2 \implies in~relatia~(b+12)(c-12)=-144,~ar~rezulta\\ca~10~divide~pe~144,~fals,~acest~caz~cade.\\\boxed{\it daca~c=3} \implies 4+a=16 \implies \boxed{\it a=12},~de~unde~se\\obtine~\boxed{\it b=4}~.$