Question

\(\left\{{}\begin{matrix}xy+z=2014\\x+yz=2013\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}xy+z=2014\left(1\right)\\x+yz=2013\left(2\right)\end{matrix}\right.\)

Trừ theo từng vế (1) cho(2) ta được:

x(y-1)+z(1-y)=1 <=>-x(1-y)+z(1-y)=1

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

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2\end{matrix}\right.\)

Bạn thế từng th vào rồi giải ra nhé

Answer 2

sai roi,x ; co thuoc Z dau