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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`1/x+1/y=1/3(x,y in NN^**)`

`=>(x+y)/(xy)=1/3`

`=>3(x+y)=xy`

`=>3x+3y=xy`

`=>xy-3x-3y=0`

`=>x(y-3)-3(y-3)-9=0`

`=>(x-3)(y-3)=9`

Vì `x,y in NN^**=>x-3,y-3 in ZZ`

`=>x-3,y-3 in Ư(9)={+-1,+-9}`

`*x-3=-1,y-3=-9`

`=>x=2,y=-6(KTM)`

`*x-3=1,y-3=9`

`=>x=4,y=12(tm)`

`*y-3=-1,x-3=-9`

`=>y=2,x=-6(KTM)`

`*y-3=1,x-3=9`

`=>y=4,x=12(tm)`

Vậy `(x,y)=(4,12),(12,4)`

Tu de bai suy ra 2y+2x=xy<=>...<=>y(2-x)= -2x<=>y=2x/(x-2)<=>y=(2x-4+4)/(x-2)<=>y=2+4/(x-2)

vi x la so nguyen Dưỡng nen x-2 la so nguyen  duong va la ước cua 4 => x-2 =1 hoặc x-2= 4 => x=3 hoac x=6 

Voi x=3 => y= 6

voi x=6=> y=3

vay cac cap so nguyen duong (x;y) can tim la (3;6); (6;3)

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Sau khi chi ra x-2 la uoc nguyen duong cua 4

 Co 3  Truong hop

x-2 =1; x-2=2;x-2=4

Tu do tinh duoc x=3;x=4;x=6. Suy ra cac gia tri tuong ung cua y

co 3 cap so nguyen duong x, y can Tim:(3;6);(4 ;4);(6;3)

a) => 2xy +3x=y+1

=> 2xy+3x-y=1

=> x(2y+3) -  1/2 (2y+3) +3/2 =1

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

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

ta có bảng

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