Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
\;5{x^2} + \;5{y^2}\;\; + \;8xy - 2x\; + \;2y\; + \;2\; = \;0\;\\
\Leftrightarrow 4{{\rm{x}}^2} + 8{\rm{x}}y + 4{y^2} + {x^2} - 2{\rm{x}} + 1 + {y^2} + 2y + 1 = 0\\
\Leftrightarrow 4\left( {{x^2} + 2{\rm{x}}y + {y^2}} \right) + \left( {{x^2} - 2{\rm{x + 1}}} \right) + \left( {{y^2} + 2y + 1} \right) = 0\\
\Leftrightarrow 4{\left( {x + y} \right)^2} + {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 0\\
Do:4{\left( {x + y} \right)^2} \ge 0\\
{\left( {x - 1} \right)^2} \ge 0\\
{\left( {y + 1} \right)^2} \ge 0\\
= > 4{\left( {x + y} \right)^2} + {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} \ge 0\\
= > \left\{ \begin{array}{l}
x + y = 0\\
x - 1 = 0\\
y + 1 = 0
\end{array} \right. = > \left\{ \begin{array}{l}
x = 1\\
y = - 1
\end{array} \right.\\
= > x + y = 1 - 1 = 0\\
M = {\left( {\;{\rm{x}} + {\rm{y}}\;} \right)^{2007}}\; + \;{\left( {\;{\rm{x}} - 2\;} \right)^{2008}}\; + \,{\left( {\;{\rm{y}} + 1\;} \right)^{2009}}\\
= {0^{2007}} + {\left( {1 - 2} \right)^{2008}} + {\left( { - 1 + 1} \right)^{2009}}\\
= 0 + {( - 1)^{2008}} + {0^{2009}}\\
= 0 + 1 + 0\\
= 1\\
vay:M = 1
\end{array}\]

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Answer 2

\[\begin{array}{l}
\;5{x^2} + \;5{y^2}\;\; + \;8xy - 2x\; + \;2y\; + \;2\; = \;0\;\\
\Leftrightarrow 4{{\rm{x}}^2} + 8{\rm{x}}y + 4{y^2} + {x^2} - 2{\rm{x}} + 1 + {y^2} + 2y + 1 = 0\\
\Leftrightarrow 4\left( {{x^2} + 2{\rm{x}}y + {y^2}} \right) + \left( {{x^2} - 2{\rm{x + 1}}} \right) + \left( {{y^2} + 2y + 1} \right) = 0\\
\Leftrightarrow 4{\left( {x + y} \right)^2} + {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 0\\
Do:4{\left( {x + y} \right)^2} \ge 0\\
{\left( {x - 1} \right)^2} \ge 0\\
{\left( {y + 1} \right)^2} \ge 0\\
= > 4{\left( {x + y} \right)^2} + {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} \ge 0\\
= > \left\{ \begin{array}{l}
x + y = 0\\
x - 1 = 0\\
y + 1 = 0
\end{array} \right. = > \left\{ \begin{array}{l}
x = 1\\
y = - 1
\end{array} \right.\\
= > x + y = 1 - 1 = 0\\
M = {\left( {\;{\rm{x}} + {\rm{y}}\;} \right)^{2007}}\; + \;{\left( {\;{\rm{x}} - 2\;} \right)^{2008}}\; + \,{\left( {\;{\rm{y}} + 1\;} \right)^{2009}}\\
= {0^{2007}} + {\left( {1 - 2} \right)^{2008}} + {\left( { - 1 + 1} \right)^{2009}}\\
= 0 + {( - 1)^{2008}} + {0^{2009}}\\
= 0 + 1 + 0\\
= 1\\
vay:M = 1
\end{array}\]

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