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 $x=1;y=-2$