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Posted: **Tue Nov 25, 2008 10:58 pm**

1. Daca \( \frac{1}{a+2001}+\frac{1}{2b+2001}+\frac{1}{3c+2001}=\frac{3}{2002} \), calculati \( \frac{a}{a+2001}+\frac{2b}{2b+2001}+\frac{3c}{3c+2001} \).

2. Fie\( P={1,2,3,...,2003} \) si \( a,b,c,x,y \in P \) astfel incat:
\( \frac{a}{a+d}+\frac{b}{b+d}+\frac{c}{c+d}=x \) si \( \frac{d}{a+d}+\frac{b}{b+d}+\frac{d}{c+d}=y \). Calculati \( x \cdot y \)

Posted: **Thu Apr 23, 2009 9:56 pm**

1. \( \frac{a}{a+2001}+\frac{2b}{2b+2001}+\frac{3c}{3c+2001}=\frac{(a+2001)-2001}{a+2001}+\frac{(2b+2001)-2001}{2b+2001}+\frac{(3c+2001)-2001}{3c+2001}=1-\frac{2001}{a+2001}+1-\frac{2001}{2b+2001}+1-\frac{2001}{3c+2001}=3-2001(\frac{1}{a+2001}+\frac{1}{2b+2001}+\frac{1}{3c+2001}) \)
\( =3-2001\cdot \frac{3}{2002}=\frac{6006-6003}{2002}=\frac{3}{2002} \).

2. \( x+y=\frac{a}{a+d}+\frac{b}{b+d}+\frac{c}{c+d}+\frac{d}{a+d}+\frac{d}{b+d}+\frac{d}{c+d} \)\( =\frac{a+d}{a+d}+\frac{b+d}{b+d}+\frac{c+d}{c+d}=1+1+1=3 \).
Deoarece \( x,y \in \mathb{P} \)\( \rightarrow (x;y) \in \math {(1;2), (2;1)} \)\( \Rightarrow x\cdot y=1\cdot 2=2 \).

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