Question

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

Answer 1

\(TH_1:x-y=\sqrt{2}\left(x\ge y\right)\\ \Leftrightarrow x=\sqrt{2}+y\\ PT\left(2\right)\Leftrightarrow2\sqrt{2}+2y-y=-1\\ \Leftrightarrow y=-1-2\sqrt{2}\\ \Leftrightarrow x=\sqrt{2}-1-2\sqrt{2}=-1-\sqrt{2}\left(tm\right)\\ TH_2:y-x=\sqrt{2}\left(x< y\right)\\ \Leftrightarrow y=\sqrt{2}+x\\ PT\left(2\right)\Leftrightarrow2x-\sqrt{2}-x=-1\\ \Leftrightarrow x=\sqrt{2}-1\\ \Leftrightarrow y=\sqrt{2}+\sqrt{2}-1=2\sqrt{2}-1\left(tm\right)\)

Vậy \(\left(x;y\right)=\left(-1-\sqrt{2};-1-2\sqrt{2}\right);\left(\sqrt{2}-1;2\sqrt{2}-1\right)\)