Question

[Đỗ Minh An]

tìm x . y thuộc N sao cho $6xy - 8x - 3y - 2 = 0$

Answer 1

[Nguyễn Lê Phước Thịnh]

Ta có: \(6xy-8x-3y-2=0\)

\(\Leftrightarrow6xy-3y-8x+4-6=0\)

\(\Leftrightarrow3y\left(2x-1\right)-4\left(2x-1\right)=6\)

\(\Leftrightarrow\left(2x-1\right)\left(3y-4\right)=6\)

\(\Leftrightarrow\left(2x-1\right);\left(3y-4\right)\inƯ\left(6\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(3y-4\right)\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

mà 2x-1 lẻ và \(2x-1\ge-1\) \(\forall x\in N\)

nên \(\left(2x-1\right)\in\left\{-1;1;-3;3\right\}\) và \(\left(3y-4\right)\in\left\{2;-2;6;-6\right\}\)

Trường hợp 1: 

\(\left\{{}\begin{matrix}2x-1=-1\\3y-4=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 2: 

\(\left\{{}\begin{matrix}2x-1=1\\3y-4=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\3y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{10}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 3: 

\(\left\{{}\begin{matrix}2x-1=-3\\3y-4=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-2\\3y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 4: 

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

Vậy: (x,y)=(2;2)