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\(\Leftrightarrow\left\{{}\begin{matrix}xy-3x+2y-6=xy+1\\2x+2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y-3x=7\\2x+2y=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{29}{10}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-1\right)\left(y+3\right)=xy+27\\\left(x-2\right)\left(y+1\right)-xy=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+3x-y-3=xy+27\\xy+x-2y-2-xy=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-y=30\\x-2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-30\\x-2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-30\\x-2\left(3x-30\right)=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-30\\x-6x+60=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-30\\-5x=-50\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3.10-30\\x=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=10\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm là: \(\left\{{}\begin{matrix}y=0\\x=10\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x\left(x+1\right)=a\\y\left(y+1\right)=b\end{matrix}\right.\) thì ta có:
\(\left\{{}\begin{matrix}a+b=8\\ab=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=6\end{matrix}\right.or\left\{{}\begin{matrix}a=6\\b=2\end{matrix}\right.\)
Tới đây thì đơn giải rồi nhé
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-2y^2=2\\xy+x^2=2\end{matrix}\right.\)
Trừ vế cho vế:
\(\Rightarrow3x^2-xy-2y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(3x+2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=-\dfrac{3}{2}x\end{matrix}\right.\)
Thế vào pt đầu: \(\left[{}\begin{matrix}2x^2-x^2=1\\2x^2-\left(-\dfrac{3}{2}\right)x^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=1\\-\dfrac{1}{4}x^2=1\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=y=1\\x=y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+3x-y-3=xy+27\\xy+x-2y-2=xy+8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-y=30\\x-2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-y=30\\3x-6y=30\end{matrix}\right.\) \(\Rightarrow5y=0\Rightarrow y=0\Rightarrow x=10\)
Vậy nghiệm của hệ là \(\left(x;y\right)=\left(10;0\right)\)