Tìm x,y, biết: a) $x^{2}$$+2x+y^{2}$$+4y+5=0$ b) $x^{2}$$-4x-6y+y^{2}$$+13=0$ 03/12/2021 Bởi Charlie 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` Bình luận
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`