Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^2} + {y^2} + {z^2} = xy + yz + zx\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} = 2xy + 2yz + 2zx\\
\Leftrightarrow \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{y^2} - 2yz + {z^2}} \right) + \left( {{z^2} - 2zx + {x^2}} \right) = 0\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 0\\
{\left( {y - z} \right)^2} = 0\\
{\left( {z - x} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = y\\
y = z\\
z = x
\end{array} \right. \Leftrightarrow x = y = z\\
{x^{2014}} + {y^{2014}} + {z^{2014}} = 3\\
\Leftrightarrow {x^{2014}} + {x^{2014}} + {x^{2014}} = 3\\
\Leftrightarrow 3{x^{2014}} = 3\\
\Leftrightarrow {x^{2014}} = 1\\
\Leftrightarrow x = \pm 1\\
TH1:\,\,\,\,x = y = z = 1\\
{x^{25}} + {y^4} + {z^{2015}} = {1^{25}} + {1^4} + {1^{2015}} = 3\\
TH2:\,\,\,\,x = y = z = - 1\\
{x^{25}} + {y^4} + {z^{2015}} = {\left( { - 1} \right)^{25}} + {\left( { - 1} \right)^4} + {\left( { - 1} \right)^{2015}} = - 1 + 1 - 1 = - 1
\end{array}\)
