Question

[Thi giang Pham]

Giải hộ tớ nhá

xy-2=x+y

(x,y là số nguyên)

Answer 1

[Yeutoanhoc]

`xy-2=x+y`

`=>xy-x=y+2`

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

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

Vì `x,y in ZZ=>x,y in ZZ`

`=>x-1,y-1 in Ư(3)={1,-1,3,-3}`

Ta có bảng sau:

x-1	1	-1	3	-3
y-1	3	-3	1	-1
x	2	0	4	-2
y	4	-2	2	0
KL	TM	TM	TM	TM

Vậy `(x,y) in (2,4),(0,-2),(4,2),(-2,0)`


 

Answer 2

[꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜT๖ۣۜR๖ۣۜÂ๖ۣ...]

Giải:

\(xy-2=x+y\)  

\(\Rightarrow xy-x=y+2\) 

\(\Rightarrow x.\left(y-1\right)=y-1+3\) 

\(\Rightarrow x.\left(y-1\right)-\left(y-1\right)=3\) 

\(\Rightarrow\left(x-1\right).\left(y-1\right)=3\) 

\(\Rightarrow\left(x-1\right)\) và \(\left(y-1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\) 

Ta có bảng giá trị:

x-1	-3	-1	1	3
y-1	-1	-3	3	1
x	-2	0	2	4
y	0	-2	4	2

Vậy \(\left(x;y\right)=\left\{\left(-2;0\right);\left(0;-2\right);\left(2;4\right);\left(4;2\right)\right\}\)

Chúc bạn học tốt!