Ta có: \(a+b=1\)
\(b+c=-2\)
\(a+c=-7\)
\(\Rightarrow a+b+b+c+c+a=1+\left(-2\right)+\left(-7\right)\)
\(=2a+2b+2c=-8\)
\(\Rightarrow2\left(a+b+c\right)=-8\)
\(\Rightarrow a+b+c=-4\)
\(\Rightarrow a=-4+2=-2\)
\(\Rightarrow b=-4+7=3\)
\(\Rightarrow c=-4-1=-5\)
Vậy \(a=-2;b=3;c=-5\)
Theo đề bài ta có :
\(\left\{{}\begin{matrix}a+b=1\\b+c=-2\\a+c=-7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1-b\\c=-2-b\\1-b-2-b=-7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1-b\\c=-2-b\\b=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=1-3\\c=-2-3\\b=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=-2\\c=-5\\b=3\end{matrix}\right.\)