![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x.y+x+y=36\)
\(x\left(y+1\right)+y=36\)
\(x\left(y+1\right)+\left(y+1\right)=36+1\)
\(\left(y+1\right)\left(x+1\right)=37\)
\(\left(y+1\right)\left(x+1\right)\) có 4 cặp: \(y+1=1;x+1=37\)
\(y+1=37;x+1=1\)
\(y+1=-1;x+1=-37\)
\(y+1=-37;x+1=-1\)
\(x;y\) có 4 cặp: \(y=0;x=36\)
\(y=36;x=0\)
\(y=-2;x=-38\)
\(y=-38;x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x + y = x . y
=> xy - x - y = 0
x ( y - 1 ) - ( y -1 ) = 0 +1
( x -1 ) ( y -1 ) = 1
ta có : 1 = 1 .1 = ( -1 ) . ( -1 )
T/H1 : x -1 = 1=> x = 2
=> y - 1 = 1 = > x =2
T/H2 : x -1 = -1 => x = 0
=> y -1 = -1 => y = 0
Vậy ( x ; y ) \(\in\){ ( 2 ; 2 ) ; ( 0 ; 0 }
x+y - x+y =0
[x - xy]+y-1=-1
x.[1-y]-[-y+1]=-1
x.[1-y]-[1-y]=-1
[1-y] .[x-1]=-1
ta thay y thuoc z suy ra 1-y thuoc z
ta thay x thuoc z suy ra x-1 thuoc z
nen 1-y thuoc uoc cua -1
1-y thuoc 1 -1
ta co bang sau
1-y 1 -1
y 0 2
x-1 -1 1
x 0 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
x.y-x.1-y=x.(y-1)-y
=x.(y-1)-(y-1).1-1
=>(x-1).(y-1)-1=2
=>(x-1).(y-1)=3
=>x-1\(\in\)Ư(3)
y-1\(\in\)Ư(3)
Mà Ư(3)={-3;-1;1;3}
Ta có bảng sau:
y-1 | y | x-1 | x |
-3 | -2 | -1 | 0 |
-1 | 0 | -3 | -2 |
1 | 2 | 3 | 4 |
3 | 4 | 1 | 2 |
Vậy (x;y)\(\in\){(0;-2);(-2:0);(4:2);(2;4)}
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>(x+1)(y-1)=11
=>\(\left(x+1;y-1\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;12\right);\left(10;2\right);\left(-2;-10\right);\left(-12;0\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>x(y-1)+(y-1)=11
=>(x+1)(y-1)=11
=>\(\left(x+1;y-1\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;12\right);\left(10;2\right);\left(-2;-10\right);\left(-12;0\right)\right\}\)
Ta có:
x2y + xy - x = 6
x2y + xy - x -1 = 5
xy.(x + 1) - (x + 1) = 5
(x = 1).(xy - 1) = 1.5 = (-1).(-5) = 5.1 = (-5).(-1)
Ta có bảng giá trị;
Vậy (x;y) = (-2;2) ; (-6;0)
Cảm ơn bạn ạ!