\(x^2+2x+y^2-6y+10=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-6y+9\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-3\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{cases}}\)\(\Rightarrow\left(x+1\right)^2+\left(y-3\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
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