Question

Giải hệ phương trình:

\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)=\dfrac{1}{2}xy+50\\\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=\dfrac{1}{2}xy-32\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)=\dfrac{1}{2}xy+50\\\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=\dfrac{1}{2}xy-32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}xy+\dfrac{3}{2}x+y+3-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-x-y+2-\dfrac{1}{2}xy=-32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+y=47\left(1\right)\\-x-y=-34\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)-\left(2\right)\Leftrightarrow\dfrac{1}{2}x=13\Leftrightarrow x=26\)

Thay \(x=26\) vào phương trình (2) ta được :

\(-26-y=-34\Leftrightarrow y=8\)

Vậy \(x=26\) và \(y=8\)