Ẩn danh

`{((x-3)(y+4)=xy-4),((x-1)(y+2)=xy+6):}`

NT

\(\left\{{}\begin{matrix}\left(x-3\right)\left(y+4\right)=xy-4\\\left(x-1\right)\left(y+2\right)=xy+6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+4x-3y-12=xy-4\\xy+2x-y-2=xy+6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-3y=-4+12=8\\2x-y=6+2=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-8\\4x-3\left(2x-8\right)=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x-8\\4x-6x+24=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-8\\24-2x=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=2\cdot8-8=8\end{matrix}\right.\)

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MP
8 tháng 7 lúc 21:18

\(\left\{{}\begin{matrix}\left(x-3\right)\left(y+4\right)=xy-4\\\left(x-1\right)\left(y+2\right)=xy+6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy+4x-3y-12-xy+4=0\\xy+2x-y-2-xy-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-3y=16\\2x-y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-6y=32\\8x-4y=32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-10y=0\\8x-4y=32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\8x-4.0=32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=4\end{matrix}\right.\)

Vậy....

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