Ta có: \(5x^2+2xy+y^2-16x+16=0\)
\(\Leftrightarrow\left(4x^2-16x+16\right)+\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow4\left(x-2\right)^2+\left(x+y\right)^2=0\)
Vì \(\hept{\begin{cases}4\left(x-2\right)^2\ge0;\forall x,y\\\left(x+y\right)^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow4\left(x-2\right)^2+\left(x+y\right)^2\ge0;\forall x,y\)
Do đó \(4\left(x-2\right)^2+\left(x+y\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}4\left(x-2\right)^2=0\\\left(x+y\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\y=-2\end{cases}}\)
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