\(x^2-4x+2y-xy+9=0\)

\(\Leftrightarrow x^2-4x+4+2y-xy+5=0\)

\(\Leftrightarrow\left(x-2\right)^2-\left(x-2\right)y+5=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-y\right)=-5\)

⇒\(\left[{}\begin{matrix}\left(x-2\right)\left(x-2-y\right)=-5\cdot1\left(1\right)\\\left(x-2\right)\left(x-2-y\right)=-1\cdot5\left(2\right)\end{matrix}\right.\)

Vì đề kêu tìm nghiệm nguyên nên ta có

Th1:\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=-5\\x-2-y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=1\\x-2-y=-5\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-3\\y=-6\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=6\end{matrix}\right.\end{matrix}\right.\)

Th2:\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=-1\\x-2-y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=5\\x-2-y=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-6\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\y=6\end{matrix}\right.\end{matrix}\right.\)

Vậy .....

\(xy-2y-3x=1\)

\(\Leftrightarrow y\left(x-2\right)-3x=1\)

\(\Leftrightarrow y\left(x-2\right)-3x+6=7\)

\(\Leftrightarrow y\left(x-2\right)-3\left(x-2\right)=7\)

\(\Leftrightarrow\left(y-3\right)\left(x-2\right)=7\)

Đến đây dễ rồi bạn tự làm tiếp nhé

giup vsssssss mn

bn ơi bn lm đc bài này ko giúp mik vs

tìm x;y trong phương trình nghiệm nguyên sau:

a)x^2+y^2-2.(3x-5y)=11                b)x^2+4y^2=21+6x

a, \(xy+4x-2y=2\)

\(\Rightarrow y\left(x-2\right)+4\left(x-2\right)=-6\)

\(\Rightarrow\left(x-2\right)\left(y+4\right)=-6\)

xy+3x-2y-7=0

 =>(xy-2y)+3x-7=0

=>y(x-2)+3x-6=-1

=>y(x-2)+3(x-2)=-1

=>(y+3)(x-2)=-1

=>y+3 và x-2 thuộc Ư(1)={1;-1}

Xét y+3=1 =>y=2 <=>x-2=-1 =>x=1

Xét y+3=-1 =>y=-4 <=>x-2=1 =>x=3