Question

\(\left\{{}\begin{matrix}17x+2y=2011\left|xy\right|\\x-2y=3xy\end{matrix}\right.\)

Answer 1

- TH1: \(xy\ge0\Rightarrow\left|xy\right|=xy\)

Hệ trở thành: \(\left\{{}\begin{matrix}17x+2y=2011xy\\x-2y=3xy\end{matrix}\right.\)

Cộng vế: \(\Rightarrow18x=2014xy\Rightarrow2x\left(1007y-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\y=\dfrac{9}{1007}\Rightarrow x=\dfrac{9}{490}\end{matrix}\right.\) (đều thỏa mãn)

TH2: \(xy< 0\Rightarrow\left|xy\right|=-xy\)

Hệ trở thành: \(\left\{{}\begin{matrix}17x+2y=-2011xy\\x-2y=3xy\end{matrix}\right.\)

Cộng vế: \(18x=-2008xy\Rightarrow2x\left(9+1004y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\left(loại\right)\\y=-\dfrac{9}{1004}\Rightarrow x=-\dfrac{18}{1031}\left(loại\right)\end{matrix}\right.\)