Đáp án + Giải thích các bước giải:
$\left\{\begin{matrix}(x-1)(y-2)=(x+1)(y-3)& \\(x-5)(y+4)=(x-4)(y+1)& \end{matrix}\right.$
`<=>` $\left\{\begin{matrix}xy-y-2x+2=xy+y-3x-3& \\xy-5y+4x-20=xy-4y+x-4& \end{matrix}\right.$
`<=>` $\left\{\begin{matrix}xy-xy-y-y+3x-2x+3+2=0& \\xy-xy+4y-5y+4x-x+4-20=0& \end{matrix}\right.$
`<=>` $\left\{\begin{matrix}-2y+x+5=0& \\-y+3x-16=0& \end{matrix}\right.$
`<=>` $\left\{\begin{matrix}-2y+x=-5& \\-y+3x=16& \end{matrix}\right.$
`<=>` $\left\{\begin{matrix}-2y+x=-5& \\-2y+6x=32& \end{matrix}\right.$
`<=>` $\left\{\begin{matrix}-5x=-37& \\-y+3x=16& \end{matrix}\right.$
`<=>` $\left\{\begin{matrix}x=\dfrac{37}{5}& \\y=\dfrac{31}{5}& \end{matrix}\right.$
Vậy hệ phương trình có nghiệm là : `(x;y)=((37)/(5);(31)/(5))`
