Question

Ghpt\(\left\{{}\begin{matrix}\left|x-2\right|+2\left|y-1\right|=9\\x+\left|y-1\right|=-1\end{matrix}\right.\)

Answer 1

Nếu \(x>0\Rightarrow x+\left|y-1\right|>0>-1\) hệ vô nghiệm

\(\Rightarrow x\le0\Rightarrow\left|x-2\right|=2-x\)

Hệ trở thành: \(\left\{{}\begin{matrix}2-x+2\left|y-1\right|=9\\x+\left|y-1\right|=-1\end{matrix}\right.\)

\(\Rightarrow3\left|y-1\right|=6\Rightarrow\left|y-1\right|=2\Rightarrow\left[{}\begin{matrix}y=3\\y=-1\end{matrix}\right.\)

Thế vào \(x+\left|y-1\right|=-1\Rightarrow x\)