\(3x^2+3xy-17=7x-2y\)

\(\Leftrightarrow3x\left(x+y\right)+2x+2y-9x-17=0\)

\(\Leftrightarrow3x\left(x+y\right)+2\left(x+y\right)-9x-6-11=0\)

\(\Leftrightarrow\left(x+y\right)\left(3x+2\right)-3\left(3x+2\right)=11\)

\(\Leftrightarrow\left(3x+2\right)\left(x+y-3\right)=11\)

\(\Leftrightarrow\left(3x+2\right);\left(x+y-3\right)\in\left\{-1;1;-11;11\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-1;-7\right);\left(-\dfrac{1}{3};\dfrac{43}{3}\right);\left(-\dfrac{11}{3};\dfrac{17}{3}\right);\left(3;1\right)\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-1;-7\right);\left(3;1\right)\right\}\left(x;y\inℤ\right)\)

\(x\cdot\left(y-1\right)+y=2\\ xy-x+y=2\\ y\cdot\left(x+1\right)-x-1=2-1\\ y\cdot\left(x+1\right)-\left(x+1\right)=1\\ \left(x+1\right)\left(y-1\right)=1\)

mà `x;y in ZZ => x+1;y-1 in ZZ`

nên `x+1;y-1` thuộc ước nguyên của `1`

`=>x+1;y-1 in {1;-1}`

`=>x in {0;-2}; y in {2;0}`

\(x.\left(y-1\right)+y=2\)

\(x.\left(y-1\right)+\left(y-1\right)=2-1\)

\(\left(y-1\right)\left(x-1\right)=1\)

(y-1) ; (x-1) có 2 cặp: \(y-1=1;x-1=1\)  hoặc \(y-1=-1;x-1=-1\)

\(x;y\) có  2 cặp: \(y=2;x=2\) hoặc \(y=0;x=0\)

Ta có : \(2^{x+1}.3^y=12^x\)

\(\Leftrightarrow3^y=\dfrac{12^x}{2^{x+1}}=\dfrac{3^x.4^x}{2^{x+1}}=\dfrac{3^x.2^{2x}}{2^{x+1}}=3^x.2^{2x}:2^{x+1}=3^x.2^{x-1}\)

\(\Leftrightarrow\dfrac{3^y}{3^x}=2^{x-1}\)

\(\Leftrightarrow3^{y-x}=2^{x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-x=0\\x-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)(tm) 

Vậy (x;y) = (1;1) nghiệm của phương trình trên 

y=x=1