ta có:

x-y+y-z+x+z=8+10+12

2x=30

x=30:2=15

y=15-8=7

z=7-10=-3

khi đó: x+y+z=15+7+(-3)=19

chờ, e bít làm

nhoc quay pha làm sai rùi, thật 100%

Ta có: \(xy+12=x+y\)

\(\Leftrightarrow xy-x-y=12\)

\(\Leftrightarrow x\left(y-1\right)-\left(y-1\right)=13\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)=13=1.13=13.1=-1.-13=-13.-1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1;y-1=13\\x-1=13;y-1=1\\x-1=-1;y-1=-13\\x-1=-13;y-1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2;y=14\\x=14;y=2\\x=0;y=-12\\x=-12;y=0\end{matrix}\right.\)

Vậy:....................

xy-x-y=2

=>x(y-1)-y+1=3

=>x(y-1)-(y-1)=3

=>(y-1)(x-1)=3

lập bảng=>tìm x,y

`xy+2x+2y=-16`

`<=>x(y+2)+2y+4=-12`

`<=>x(y+2)+2(y+2)=-12`

`<=>(x+2)(y+2)=-12`

Vì `x,y in ZZ=>x+2,y+2 in ZZ`

`=>x+2,y+2 in Ư(-12)={+-1,+-2,+-3,+-4,+-6,+-12}`

Đến đấy chia th rồi giải thui :v

Vì 

xy+2x+2y+4=-16+4

x.(y+2)+2.(y+2)=-12

(x+2).(y+2)=-12

Ta có: -12=-1.12=12.-1=1.-12=-12.1=-6.2=2.-6=6.-2=-2.6=-3.4=3.-4=4.-3=-4.3

⇒(x,y)=(-3,10);(10,-3);(-1,-14);(-14,1);(1,-14);(-14,1);(4,-4),(-4,4);(-5,2);(1,-6);(2,-5);(-6,1)

=>x(y-1)+y-1=4

=>(x+1)(y-1)=4

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

=>\(\left(x,y\right)\in\left\{\left(0;5\right);\left(3;2\right);\left(-2;-3\right);\left(-5;0\right);\left(1;3\right);\left(-3;-1\right)\right\}\)

Ta có :xy-y=3

=>y(x-1)=3

Vậy y;x-1 \(\inƯ\left(3\right)\)

Ta có bảng sau:

Vậy với ..................

=> x( y - 2 ) = 7

=> x ; ( y -2 ) thuộc Ư(7) = -7 ; -1 ; 1 ; 7

Th1 x = -7

       y - 2 = -1 => y = -1 + 2 = 1

Th2  x = -1

        y - 2 = -7 => y = -7 + 2 = -5

Th3  x = 1

        y - 2 = 7 => y = 7 + 2 = 9

Th4  x = 7

        y - 2 = 1 => y = 1 + 2 = 3