a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

\(A=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

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`xy-2x+y+1=0`

`x(y-2)+(y-2)+3=0`

`(y-2)(x+1)=-3=-3.1=-1.3`

`@{(x+1=-3),(y-2=1):}=>{(x=-4),(y=3):}`

`@{(x+1=3),(y-2=-1):}=>{(x=2),(y=1):}`

`@{(x+1=-1),(y-2=3):}=>{(x=-2),(y=5):}`

`@{(x+1=1),(y-2=-3):}=>{(x=0),(y=-1):}`

Vậy cặp x;y =2;2

xy=x+y 
=> x(y-1)=y (*) 
=> x=y/(y-1) 
Để x nguyên thì y chia hết cho y-1 
do y, y-1 luôn nguyên tố cùng nhau với y-1>=2 hoặc y-1<=-2 
=> y-1=1 hoặc y-1=-1 
TH1: Nếu y-1=1 
=>y=2 
(*) => x=2 

TH2 :Nếu y-1=-1 => y=0 và x=0 

Vậy có cặp số nguyên (x;y) =(2,2) và (0,0).

=>  xy-(x+y)=0

=>xy-x+y = 0

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

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

a) x + y +xy = 6

y( 1 + x ) + x + 1 = 7

( x + 1 ) ( y + 1 ) = 7

b) 2x + y - 2xy - 8 = 0

2x ( 1 - y ) - ( 1 - y ) - 7 = 0

( 1 - y ) ( 2x - 1 ) = 7

c) x - 4y + xy - 1 = 0

x( 1 + y ) -4( 1 + y ) + 3 = 0

( 1 + y ) ( x- 4 ) = 3

\(\Leftrightarrow2x^2-x+1=xy+2y\)

\(\Leftrightarrow2x^2-x+1=y\left(x+2\right)\)

\(\Leftrightarrow y=\dfrac{2x^2-x+1}{x+2}=2x-5+\dfrac{11}{x+2}\)

Do y nguyên \(\Rightarrow\dfrac{11}{x+2}\) nguyên \(\Rightarrow x+2=Ư\left(11\right)\)

Mà x nguyên dương \(\Rightarrow x+2\ge3\Rightarrow x+2=11\Rightarrow x=9\)

\(\Rightarrow y=14\)

Vậy \(\left(x;y\right)=\left(9;14\right)\)

4:

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

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

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