Tìm x,y, biết: a) $x^{2}$$+2x+y^{2}$$+4y+5=0$ b) $x^{2}$$-4x-6y+y^{2}$$+13=0$

a.

`x^2+2x+y^2+4y+5=0`

`⇒x^2+2x+y^2+4y+1+4=0`

`⇒(x^2+2x+1)+(y^2+4y+4)=0`

`⇒(x+1)^2+(y+2)^2=0`

Do $\begin{cases}(x+1)^2≥0\\(y+2)^2≥0\end{cases}$

`⇒(x+1)^2+(y+2)^2≥0`

Mà `(x+1)^2+(y+2)^2=0`

`⇒` $\begin{cases}(x+1)^2=0\\(y+2)^2=0\end{cases}$

`⇒` $\begin{cases}x+1=0\\y+2=0\end{cases}$

`⇒` $\begin{cases}x=-1\\y=-2\end{cases}$

Vậy `x=-1; y=-2`

b.

`x^2-4x-6y+y^2+13=0`

`⇒x^2-4x-6y+y^2+4+9=0`

`⇒(x^2-4x+4)+(y^2-6y+9)=0`

`⇒(x-2)^2+(y-3)^2=0`

Do $\begin{cases}(x-2)^2≥0\\(y-3)^2≥0\end{cases}$

`(x-2)^2+(y-3)^2≥0`

Mà `(x-2)^2+(y-3)^2=0`

`⇒` $\begin{cases}(x-2)^2=0\\(y-3)^2=0\end{cases}$

`⇒` $\begin{cases}x-2=0\\y-3=0\end{cases}$

`⇒` $\begin{cases}x=2\\y=3\end{cases}$

Vậy `x=2; y=3`