Chủ đề này có 1 trả lời

[Mr0]

$\left\{\begin{matrix}
x+y=2\\
x^{2012}+y^{2012}=x^{2011}+y^{2011}
\end{matrix}\right.$

[Crystal]

$\left\{\begin{matrix}
x+y=2\\
x^{2012}+y^{2012}=x^{2011}+y^{2011}
\end{matrix}\right.$

Hệ phương trình tương đương với: $$\left\{ \begin{gathered}
x + y = 2 \\
2\left( {{x^{2012}} + {y^{2012}}} \right) = 2\left( {{x^{2011}} + {y^{2011}}} \right) \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x + y = 2 \\
2\left( {{x^{2012}} + {y^{2012}}} \right) = \left( {x + y} \right)\left( {{x^{2011}} + {y^{2011}}} \right) \\
\end{gathered} \right.$$

$$ \Leftrightarrow \left\{ \begin{array}{l}
x + y = 2\\
2{x^{2012}} + 2{y^{2012}} = {x^{2012}} + x{y^{2011}} + {x^{2011}}y + {y^{2012}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + y = 2\\
{x^{2012}} + {y^{2012}} = x{y^{2011}} + {x^{2011}}y
\end{array} \right.$$

$$ \Leftrightarrow \left\{ \begin{array}{l}
x + y = 2\\
{x^{2011}}\left( {x - y} \right) - {y^{2011}}\left( {x - y} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + y = 2\\
\left( {x - y} \right)\left( {{x^{2011}} - {y^{2011}}} \right) = 0
\end{array} \right.$$

$$ \Leftrightarrow \left\{ \begin{array}{l}
x + y = 2\\
x = y
\end{array} \right. \Leftrightarrow x = y = 1$$