Theo đề bài, ta có:
\(\left[\left(a+b\right)+\left(b+c\right)\right]\div2=15\) \(\Rightarrow\left(a+b\right)+\left(b+c\right)=30\) \(\Rightarrow a+2b+c=30\)
\(\left[\left(b+c\right)+\left(c+a\right)\right]\div2=10\) \(\Rightarrow\left(b+c\right)+\left(c+a\right)=20\) \(\Rightarrow a+b+2c=20\)
\(\left[\left(a+c\right)+\left(a+b\right)\right]\div2=11\) \(\Rightarrow\left(a+c\right)+\left(a+b\right)=22\) \(\Rightarrow2a+b+c=22\)
Cộng vế với vế, ta có:
\(\left(a+2b+c\right)+\left(a+b+2c\right)+\left(2a+b+c\right)=30+20+22\) \(\Rightarrow4a+4b+4c=72\) \(\Rightarrow4\times\left(a+b+c\right)=72\) \(\Rightarrow a+b+c=18\)
\(\Rightarrow\left\{{}\begin{matrix}a=\left(2a+b+c\right)-\left(a+b+c\right)\\b=\left(a+2b+c\right)-\left(a+b+c\right)\\c=\left(a+b+2c\right)-\left(a+b+c\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=22-18\\b=30-18\\c=20-18\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=4\\b=12\\c=2\end{matrix}\right.\)
Vậy \(a=4,b=12\) và \(c=2\).
