=>xy+x+y=12

=>y(x+1)+x+1=13

=>(x+1)(y+1)=13

=>\(\left(x+1;y+1\right)\in\left\{\left(1;13\right);\left(13;1\right);\left(-1;-13\right);\left(-13;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;12\right);\left(12;0\right);\left(-2;-14\right);\left(-14;-2\right)\right\}\)

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

\(x^2-y^2=12\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=12\)

mà \(x,y\in Z\)

\(\Rightarrow x-y,x+y\in\)ước của \(12\)

làm nốt nha

4:

(x+1)(y-2)=5

=>\(\left(x+1;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;7\right);\left(4;3\right);\left(-2;-3\right);\left(-6;1\right)\right\}\)

Ta có: 2^x-1.3^y+1=12^x+y

=>2^x-1.3^y+1=(4.3)^x+y

=>2^x-1.3^y+1=4^x+y.3^x+y

=>4^x+y:2^x-1=3^y+1:3^x+y

=>2^2x+2y:2^x-1=3^y+1:3^x+y

=>2^2x+2y-x+1=3^y+1-x-y

=>2^x+2y+1=3^1-x

*Xét x+2y+1=0=>2^x+2y+1=2⁰=1=3^1-x=3⁰=>1-x=0=>x=1

=>x+2y+1=0=>1+2y+1=0=>2+2y=0=>2y=-2=>y=-1

*Xét x+2y+1>0=>2^x+2y+1 chia hết cho 2=>3^1-x chia hết cho 2

=>Vô lí

Vậy x=1,y=-1

neu la so nguyen duong con tim duoc chu tinh so nguyen am thi chiu

2^x-1 . 3^y+1 = 12^x+y

2^x-1 . 3^y+1 = 3^x+y . 2^2(x+y)

y + 1 = x + y => x = 1

x - 1 = 2(x + y)

2x + 2y = 1 - 1 = 0

2.1 + 2.y = 0

2 + 2y = 0

2y = -2 => y = -1

Vậy x = 1 ; y = -1