Đáp án: $a-b+c=8$
Giải thích các bước giải:
$\begin{array}{l}
+ ){a^2} + {b^2} + {c^2} = {\left( {a + b} \right)^2} + {\left( {b + c} \right)^2} + {\left( {a + c} \right)^2}\\
\Leftrightarrow {a^2} + {b^2} + {c^2} = 2\left( {{a^2} + {b^2} + {c^2}} \right) + 2\left( {ab + bc + ac} \right)\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) = 0\\
\Leftrightarrow {\left( {a + b + c} \right)^2} = 0 \Leftrightarrow a + b + c = 0(1)\\
+ )ca - ab - bc = 16(2)\\
\Leftrightarrow ca + c\left( { - a - b} \right) = 16\\
\Leftrightarrow ca + {c^2} = 16\\
\Leftrightarrow c\left( {c + a} \right) = 16\\
\Leftrightarrow - bc = 16(3) \Rightarrow b < 0\left( {do{\rm{ }}c > 0} \right)\\
+ )(2),(3) \Rightarrow ca - ab = 0\\
\Leftrightarrow a\left( {c - b} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a = 0\\
c = b\left( {vl} \right)
\end{array} \right. \Leftrightarrow a = 0\\
(1) \Rightarrow b + c = 0 \Rightarrow b = - c{\rm{ }} \Rightarrow {c^2} = 16 \Rightarrow c = 4 \Rightarrow b = - 4\\
\Rightarrow a - b + c = 0 - \left( { - 4} \right) + 4 = 8
\end{array}$
Vậy $a-b+c=8$
