Question

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

Answer 1

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

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

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

Answer 2

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!

Answer 3

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

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

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