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\(3xy-4x+2y=1\Rightarrow x\left(3y-4\right)=1-2y\Rightarrow x=\dfrac{1-2y}{3y-4}\)
-Vì x,y nguyên nên \(\left(1-2y\right)⋮\left(3y-4\right)\)
\(\Rightarrow\left(3-6y\right)⋮\left(3y-4\right)\)
\(\Rightarrow\left(-6y+8-5\right)⋮\left(3y-4\right)\)
\(\Rightarrow-5⋮\left(3y-4\right)\)
\(\Rightarrow3y-4\inƯ\left\{-5\right\}\)
\(\Rightarrow3y-4\in\left\{1;5;-1;-5\right\}\)
\(\Rightarrow y\in\left\{3;1\right\}\)
*\(y=1\Rightarrow x=\dfrac{1-2.1}{3.1-4}=1\)
*\(y=3\Rightarrow x==\dfrac{1-2.3}{3.3-4}=-1\)
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\(3xy-3x-y=0\)
\(\Rightarrow3x\left(y-1\right)-\left(y-1\right)=0+1\)
\(\Rightarrow\left(3x-1\right)\left(y-1\right)=1\)
\(\Rightarrow\left(3x-1\right);\left(y-1\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)
Xét bảng
3x-1 | 1 | -1 |
y-1 | 1 | -1 |
x | 2/3 | 0 |
y | 2 | 0 |
Vậy.............................
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