a, x=1; y=2 => 12

x=2; y=1 => 21

b, x=1; y=5 => 15

x=5; y=1 => 51

c, x=1; y=6 => 16

x=6;y=1 => 61

x=2; y=3=> 23

x=3; y=2 => 32

d, x=1; y=8 => 18

x=2; y=4 => 24

x=4; y=2 => 42

x=8; y=1 => 81

5, 

x=3; y=4 => 34

x=4; y=3 => 43

x=2; y=6 => 26

x=6; y=2 => 62

`(xy-5)(xy+2)+3(xy-2)(xy+2)-(3xy-1/2)^2 +5x^2y^2`

`=(xy+2)(xy-5+3xy-6)-(9x^2y^2-3xy+1/4)+5x^2y^2`

`=(xy+2)(4xy-11)-9x^2y^2+3xy-1/4+5x^2y^2`

`=4x^2y^2-11xy+8xy-22-4x^2y^2+3xy-1/4`

`=-89/4`

Ta có: \(\left(xy-5\right)\left(xy+2\right)+3\left(xy-2\right)\left(xy+2\right)-\left(3xy-\dfrac{1}{2}\right)^2+5x^2y^2\)

\(=x^2y^2-3xy-10+3x^2y^2-12-\left(9x^2y^2-3xy+\dfrac{1}{4}\right)+5x^2y^2\)

\(=9x^2y^2-3xy-22-9x^2y^2+3xy-\dfrac{1}{4}\)

\(=\dfrac{-89}{4}\)

B-(\(3x^6-4xy^5+\dfrac{1}{3}xy^2\))=

B= \(\left(7x^6-\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}\right)+\left(3x^6-4xy^5+\dfrac{1}{3}xy^2-\dfrac{3}{2}\right)\)

B= \(7x^6-\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}+3x^6-4xy^5+\dfrac{1}{3}xy^2-\dfrac{3}{2}\)

B= \(7x^6+3x^6-\dfrac{1}{2}xy^5-4xy^5-xy^2+\dfrac{1}{3}xy^2-\dfrac{1}{3}+\dfrac{2}{3}\)

B= \(10x^6-\dfrac{9}{2}xy^5-\dfrac{2}{3}xy^2+\dfrac{1}{3}\)

1. \(A+7x^2y-5xy^2-xy=x^2y+8xy^2-5xy\)

\(\Rightarrow A+7x^2y-x^2y-5xy^2-8xy^2-xy+5xy=0\)

\(\Rightarrow A+6x^2y-13xy^2+4xy=0\)

\(\Rightarrow A=-6x^2y+13xy^2-4xy\)

2. \(4xy^2-7x+1-A=3x^2-7x-1\)

\(\Rightarrow4xy^2-3x^2-7x+7x+1+1-A=0\)

\(\Rightarrow4xy^2-3x^2+2-A=0\)

\(\Rightarrow A=4xy^2-3x^2+2\)

\(\begin{array}{l}T + H = 3{x^2}y - 2x{y^2} + xy + \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy - 2{x^2}y + 3x{y^2} + 1\\ = \left( {3{x^2}y - 2{x^2}y} \right) + \left( { - 2x{y^2} + 3x{y^2}} \right) + xy + 1\\ = {x^2}y + x{y^2} + xy + 1\\T - H = 3{x^2}y - 2x{y^2} + xy - \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy + 2{x^2}y - 3x{y^2} - 1\\ = \left( {3{x^2}y + 2{x^2}y} \right) + \left( { - 2x{y^2} - 3x{y^2}} \right) + xy - 1\\ = 5{x^2}y - 5x{y^2} + xy - 1\end{array}\)

Chọn B.

a) xy-x-y= -1

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

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

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

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

=> +) y-1=0  => y=1

Hoặc

    +) x-1=0   => x=1

d,xy+y=5

=xy+y1=5

=y[x+1]=5

Suy ra y và x+1 thuộc ước của 5 

     tíc dùm nha

      tíc 

bằng 831 giúp mình giải bài này

​

Xy6+xy

a Ta có

xy -x-y=-1

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

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

=> + y-1 =0 và x-1 thỏa mãn với mọi số nguyên

   + x-1=0 và y-1 thỏa mãn với mọi số nguyên 

nhưng tớ muốn làm hết luôn cơ !

(xy – 1)(xy + 5)

= xy.(xy + 5) + (–1).(xy + 5)

= xy.xy + xy.5 + (–1).xy + (–1).5

= x2y2 + 5xy – xy – 5

= x2y2+ 4xy – 5.