\(\left(x+y\right)^2+2\left|y-1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Giải:
\(\left(x+y\right)^2+2.\left|y-1\right|=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+y\right)^2=0\\2.\left|y-1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Chúc bạn học tốt!
\(\left(x+y\right)\)\(^2\)\(+2\left|y-1\right|\)\(=0\)
\(=>\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)