1. Ta có: \(\left|12-3x\right|-20=10\)

\(\Rightarrow\left|12-3x\right|=30\)

\(\Rightarrow12-3x=\pm30\)

Nếu \(12-3x=30\Rightarrow3x=-18\Rightarrow x=-6\)

Nếu \(12-3x=-30\Rightarrow3x=42\Rightarrow x=14\)

2. Ta có: \(\left(x+1\right)\left(y-2\right)=13=1.13=13.1=\left(-13\right).\left(-1\right)=\left(-1\right)\left(-13\right)\)

a: x-y+xy-9=0

=>x+xy-y-1=8

=>(y+1)(x-1)=8

=>(x-1;y+1) thuộc {(1;8); (8;1); (-1;-8); (-8;-1); (2;4); (4;2); (-2;-4); (-4;-2)}

=>(x,y) thuộc {(2;7); (9;0); (0;-9); (-7;-2); (3;3); (5;1); (-1;-5); (-3;-3)}

b: xy-3y-5x+10=0

=>y(x-3)-5x+15=5

=>(x-3)(y-5)=5

=>(x-3;y-5) thuộc {(1;5); (5;1); (-1;-5); (-5;-1)}

=>(x,y) thuộc {(4;10); (8;6); (2;0); (-2;4)}

c: 6xy-3x-2y-1=0

=>3x(2y-1)-2y+1-2=0

=>(2y-1)(3x-1)=2

=>(3x-1;2y-1) thuộc {(2;1); (-2;-1)}

=>(x,y) thuộc {(1;1)}

ko biet

Bài 1: 

Để E nguyên thì \(x+5⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{3;1;9;-5\right\}\)

Thank you.

giúp mình với ạ cần luôn nhá. mk sẽ tick cho!

\(xy+x-y=x\left(y+1\right)-y=4\)
\(x\left(y+1\right)-y-1=3\)
\(x\left(y+1\right)-\left(y+1\right)=3\)
\(\left(x-1\right)\left(y+1\right)=3\)
\(=>\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.=>\left[{}\begin{matrix}y+1=3\\y+1=-3\\y+1=1\\y+1=-1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.=>\left[{}\begin{matrix}y=2\\y=-4\\y=0\\y=-2\end{matrix}\right.\)

\(\text{x(y+1)-y=4}\)

\(\text{x(y+1)-y-1=3}\)

\(\text{x(y+1)-(y+1)=3}\)

\(\text{(x-1)(y+1)=3}\)

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

==> bảng:

=>  (x;y) ∈ { (2;2);(4;0);(0;-4);(-2;-2)}

<=> y(x-1) + x-1 =0 <=> (y+1)(x-1)=0 do y thuộc N nên y+1> =>x=1 

vậy (x;y)=(1;p)  (trong đó p thuộc N)