a) \(\left(2+x\right)\left(-1+2y\right)=5\)
Từ đó ta có bảng sau:

 b) \(3xy-y+3x=5\)
\(\Rightarrow y\left(3x-1\right)+\left(3x-1\right)=4\)
\(\Rightarrow\left(y+1\right)\left(3x-1\right)=4\)
Từ đó ta có bảng sau:

3xy+2y=2-x

=>3xy+2y+x=2

=>\(y\left(3x+2\right)+x+\dfrac{2}{3}=2+\dfrac{2}{3}=\dfrac{8}{3}\)

=>\(3y\left(x+\dfrac{2}{3}\right)+\left(x+\dfrac{2}{3}\right)=\dfrac{8}{3}\)

=>\(\left(x+\dfrac{2}{3}\right)\left(3y+1\right)=\dfrac{8}{3}\)

=>\(\left(3x+2\right)\left(3y+1\right)=8\)

=>\(\left(3x+2;3y+1\right)\in\left\{\left(1;8\right);\left(8;1\right);\left(-1;-8\right);\left(-8;-1\right);\left(2;4\right);\left(4;2\right);\left(-2;-4\right);\left(-4;-2\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-\dfrac{1}{3};\dfrac{7}{3}\right);\left(2;0\right);\left(-1;-3\right);\left(-\dfrac{10}{3};-\dfrac{2}{3}\right);\left(0;1\right);\left(\dfrac{2}{3};\dfrac{1}{3}\right);\left(-\dfrac{4}{3};-\dfrac{5}{3}\right);\left(-2;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;0\right);\left(-1;-3\right);\left(0;1\right);\left(-2;-1\right)\right\}\)

Ta có:3xy-5=x\(^2\)+2y

\(\Rightarrow\)3xy-2y=x\(^2\)+5   (1)

Vì x,y là số nguyên nên:x\(^2\)+5 chia hết cho 3x-2

=>9(x^2+5) chia hết cho 3x-2

9x^2+45 chia hết cho3y-2

=>9x^2-6x+6x-4+49 chia hêt cho 3x-2

=>3x(3x-2)+2(3x-2)+49 chia hết cho 3x-2

=>46 chia hết cho 3x-2

=>3x-2\(\in\)(49;-49;7;-7;1;-1)

<=>3x\(\in\)(51;-47;9;-5;3;1)

<=>x\(\in\)(1;3;17)

Thay x lần lượt vào (1) ta được y=6 hoặc y=2

Vậy y=2 hoặc y=2

Tích đúng nha!Hì hì...

Ths pn chá!!!

Ta có:3xy-5=x2+2y

⇒3xy-2y=x2+5   (1)

Vì x,y là số nguyên nên:x2+5 chia hết cho 3x-2

=>9(x^2+5) chia hết cho 3x-2

9x^2+45 chia hết cho3y-2

=>9x^2-6x+6x-4+49 chia hêt cho 3x-2

=>3x(3x-2)+2(3x-2)+49 chia hết cho 3x-2

=>46 chia hết cho 3x-2

=>3x-2∈(49;-49;7;-7;1;-1)

<=>3x∈(51;-47;9;-5;3;1)

<=>x

\(3xy-5=x^2+2y\)

\(3xy-5-x^2+2y=0\)

đến đây bn giải hệ pt bậc 2 là đc 

Vãi cả hệ pt bậc hai

\(3xy-5=x^2+2y\)

\(\Leftrightarrow3xy-2y=x^2+5\)

\(\Leftrightarrow y\left(3x-2\right)=x^2+5\)

\(\Rightarrow x^2+5⋮3x-2\)

\(\Rightarrow9\left(x^2+5\right)⋮3x-2\)

\(\Rightarrow9x^2+45⋮3x-2\)

\(\Rightarrow9x^2-6x+6x-4+49⋮3x-2\)

\(\Rightarrow3x\left(3x-2\right)+2\left(3x-2\right)+49⋮3x-2\)

Mà \(3x\left(3x-2\right)⋮3x-2\)và \(2\left(3x-2\right)⋮3x-2\)

nên \(49⋮3x-2\)

Để ý 3x - 2 chia 3 dư 1 và x nguyên nên \(3x-2\in\left\{49;7;1\right\}\)

Xét từng trường hợp, ta được: \(x\in\left\{17;3;1\right\}\)

Thay vào tính y...

\(x^2+2y^2+3xy=5\)

=>\(x^2+xy+2xy+2y^2=5\)

=>\(x\left(x+y\right)+2y\left(x+y\right)=5\)

=>\(\left(x+y\right)\left(x+2y\right)=5\)

=>\(\left(x+y\right)\left(x+2y\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

TH1: \(\left\{{}\begin{matrix}x+y=1\\x+2y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y-x-2y=1-5=-4\\x+y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-y=-4\\x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=1-y=1-4=-3\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=5\\x+2y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y-x-2y=5-1\\x+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-y=4\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-4\\x=5-y=5-\left(-4\right)=9\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}x+y=-1\\x+2y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y-x-2y=-1-\left(-5\right)\\x+2y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-y=-1+5=4\\x+2y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-4\\x=-5-2y=-5-2\cdot\left(-4\right)=-5+8=3\end{matrix}\right.\)

TH4: \(\left\{{}\begin{matrix}x+y=-5\\x+2y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y-x-2y=-5-\left(-1\right)\\x+y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-y=-5+1=-4\\x+y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\\x=-5-y=-5-4=-9\end{matrix}\right.\)