Question

Tìm x y

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

Answer 1

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

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\(y+2)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

vậy \(x=1\)

\(y=-2\)

Answer 2

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

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\\left(y+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy, ....