\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
vì: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)
=> \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
=> Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy:........................
Ta có :
\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
\(\left(y+2\right)^2\ge0\forall y\)
Nên : \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(x=1;y=-2\)
