$\text{Đáp án + Giải thích các bước giải:}$
$\text{Bổ sung đề : Tìm x ; y ∈ Z}$
`2x(3y-2)+(3y-2)=-15`
`=>(3y-2)(2x+1)=-15`
$\text{Mà x ; y ∈ Z}$
`=>(3y-2)(2x+1)=-15=(-1).15=1.(-15)=(-3).5=3.(-5)`
$\text{Lập bảng , ta có :}$
$\begin{array}{|c|c|}\hline 2x+1&-1&15&1&-15&-3&5&3&-5\\\hline 3y-2&15&-1&-15&1&5&-3&-5&3\\\hline\end{array}$
`⇒`
$\begin{array}{|c|c|}\hline x&-1(TM)&7(TM)&0(TM)&-8(TM)&-2(TM)&2(TM)&1(TM)&-3(TM)\\\hline y&\dfrac{17}{3}(KTM)&\dfrac{1}{3}(KTM)&-\dfrac{13}{3}(KTM)&1(TM)&\dfrac{7}{3}(KTM)&-\frac{1}{3}(KTM)&-1(TM)&\dfrac{5}{3}(KTM)\\\hline\end{array}$
$\text{Vậy}$ `(x;y)=(-8;1);(1;-1)`
