a.
\(A\in Z\Rightarrow2A\in Z\Rightarrow\dfrac{2\left(3x-2\right)}{2x-3}\in Z\)
\(\Rightarrow\dfrac{6x-4}{2x-3}\in Z\Rightarrow\dfrac{3\left(2x-3\right)+5}{2x-3}\in Z\)
\(\Rightarrow3+\dfrac{5}{2x-3}\in Z\Rightarrow\dfrac{5}{2x-3}\in Z\)
\(\Rightarrow2x-3=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x=\left\{-1;1;2;4\right\}\)
Thử lại thấy đều thỏa mãn
b.
\(\dfrac{2x+4}{y}-\dfrac{2}{x}-\dfrac{5}{xy}=1\)
\(\Rightarrow x\left(2x+4\right)-2y-5=xy\)
\(\Leftrightarrow2x^2+4x-y-5=xy\)
\(\Leftrightarrow2x^2+4x-5=y\left(x+1\right)\)
Với \(x=-1\) không thỏa mãn
Với \(x\ne-1\Rightarrow y=\dfrac{2x^2+4x-5}{x+1}\) (1)
Do \(y\in Z\Rightarrow\dfrac{2x^2+4x-5}{x+1}\in Z\Rightarrow\dfrac{2x^2+4x+2-7}{x+1}\in Z\)
\(\Rightarrow\dfrac{2\left(x+1\right)^2-7}{x+1}\in Z\Rightarrow2\left(x+1\right)-\dfrac{7}{x+1}\in Z\)
\(\Rightarrow\dfrac{7}{x+1}\in Z\) do \(2\left(x+1\right)\in Z\) với \(x\in Z\)
\(\Rightarrow x+1=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x=-8\left(loại\right)\\x=-2\left(loại\right)\\x=0\\x=6\end{matrix}\right.\)
Thế vào (1): \(\left[{}\begin{matrix}x=0\Rightarrow y=-5< 0\left(loại\right)\\x=6\Rightarrow y=13\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;13\right)\)
