b) Ta có: \(\left(2x+1\right)\left(3y-2\right)=12\)
\(\Leftrightarrow2x+1\) và 3y-2 là các ước của 12
Trường hợp 1:
\(\left\{{}\begin{matrix}2x+1=1\\3y-2=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{14}{3}\end{matrix}\right.\)(loại)
Trường hợp 2:
\(\left\{{}\begin{matrix}2x+1=12\\3y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=11\\3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{2}\\y=1\end{matrix}\right.\)(loại)
Trường hợp 3:
\(\left\{{}\begin{matrix}2x+1=2\\3y-2=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\3y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{8}{3}\end{matrix}\right.\)(loại)
Trường hợp 4:
\(\left\{{}\begin{matrix}2x+1=6\\3y-2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{4}{3}\end{matrix}\right.\)(loại)
Trường hợp 5:
\(\left\{{}\begin{matrix}2x+1=3\\3y-2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)(nhận)
Trường hợp 6:
\(\left\{{}\begin{matrix}2x+1=4\\3y-2=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=3\\3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{5}{3}\end{matrix}\right.\)(loại)
Trường hợp 7:
\(\left\{{}\begin{matrix}2x+1=-1\\3y-2=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-2\\3y=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{-10}{3}\end{matrix}\right.\)(loại)
Trường hợp 8:
\(\left\{{}\begin{matrix}2x+1=-12\\3y-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-13\\3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-13}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)(loại)
Trường hợp 9:
\(\left\{{}\begin{matrix}2x+1=-2\\3y-2=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-3\\3y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{2}\\y=\dfrac{-4}{3}\end{matrix}\right.\)(loại)
Trường hợp 10:
\(\left\{{}\begin{matrix}2x+1=-6\\3y-2=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-7\\3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\y=0\end{matrix}\right.\)(loại)
Trường hợp 11:
\(\left\{{}\begin{matrix}2x+1=-3\\3y-2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-4\\3y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-\dfrac{2}{3}\end{matrix}\right.\)(loại)
Trường hợp 12:
\(\left\{{}\begin{matrix}2x+1=-4\\3y-2=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-5\\3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{2}\\y=\dfrac{-1}{3}\end{matrix}\right.\)(loại)
Vậy: (x,y)=(1;2)
Lời giải phần a:
a)
$3xy+9x-2y=10$
$\Leftrightarrow 3x(y+3)-2(y+3)=4$
$\Leftrightarrow (3x-2)(y+3)=4$
Đến đây, do $3x-2,y+3$ đều là số nguyên, $3x-2$ chia $3$ dư $1$ nên ta xét các TH sau:
$3x-2=1; y+3=4\Rightarrow x=1; y=-1$
$3x-2=4; y+3=1\Rightarrow x=2; y=-2$
$3x-2=-2; y+3=-2\Rightarrow x=0; y=-5$
