Question

[Diệu Anh Hoàng]

Tìm x, y biết: \(x^2+y^2-2x+4y+5=0\)

Answer 1

[Asuka]

(x²-2x+1²)+(y²+2.2y+2²)=0

(x-1)²+(y+2)²=0

=>x-1=0=>x=1

=>y+2=0=>y=-2

Answer 2

[kudo shinichi]

\(x^2+y^2-2x+4y+5=0\)

\(\left(x^2-2x+1\right)+\left(y^2+2.y.2+2^2\right)=0\)

\(\left(x-1\right)^2+\left(y+2\right)^2=0\)

Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x;y\)

Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)

Vậy \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)